On Monday, 13 November 2023 at 10:49:51 UTC+1, WM wrote:
The proof of equinumerosity by bijection between infinite sets, M and N,
is justified by mathematical induction:
Wrong, equinumerosity by bijection is a *definition* not any theorem.
A supertask is a countably infinite sequence of operations that occur
sequentially within a finite interval of time.
Wrong, a "supertask" is the *limit* of any such sequence,
and "time" is utterly irrelevant if not as an expository device.
The proof of equinumerosity by bijection between infinite sets, M and N,
is justified by mathematical induction:
A supertask is a countably infinite sequence of operations that occur sequentially within a finite interval of time.
On 13.11.2023 12:42, Julio Di Egidio wrote:
On Monday, 13 November 2023 at 10:49:51 UTC+1, WM wrote:
The proof of equinumerosity by bijection between infinite sets, M and N, >> is justified by mathematical induction:
Wrong, equinumerosity by bijection is a *definition* not any theorem.
The bijection between naturals and algebraics is defined only? It cannot
be proved?
A supertask is a countably infinite sequence of operations that occur
sequentially within a finite interval of time.
Wrong, a "supertask" is the *limit* of any such sequence,
and "time" is utterly irrelevant if not as an expository device.
A supertask is a task, not a limit.
On 13.11.2023 12:42, Julio Di Egidio wrote:
[...]
A supertask is a task, not a limit.
Look it up.
A supertask is a task, not a limit.
Look it up.
Describing and reasoning about a supertask
is a task, not a supertask,
Indexing the fractions,
not merely describing and reasoning about
indexing the fractions,
is a supertask,
On 11/13/2023 7:08 AM, WM wrote:
On 13.11.2023 12:42, Julio Di Egidio wrote:
[...]
<snip>A supertask is a task, not a limit.<https://plato.stanford.edu/Archives/sum2005/entries/spacetime-supertasks/>
Look it up.
| A supertask may be defined as
| an infinite sequence of actions or operations
| carried out in a finite interval of time.
Indexing the fractions, not merely
describing and reasoning about
indexing the fractions, is a supertask
Indexing the algebraic numbers is a supertask.
On 13.11.2023 18:15, Jim Burns wcrote:
Describing and reasoning about a supertask
is a task, not a supertask,
Indexing the fractions,
not merely describing and reasoning about
indexing the fractions,
is a supertask,
Indexing the algebraic numbers is a supertask.
Describing the method of indexing is a task.
This holds for all of Cantor's enumerations
as well as for my matrices.
If completeness is accomplished,
then it is accomplished in both cases.
However, we can _reason about_ the first,
starting with accomplishing the second,
then augmenting with not-first-false claims.
On 14.11.2023 21:44, Jim Burns wrote:
However, we can _reason about_ the first,
starting with accomplishing the second,
then augmenting with not-first-false claims.
Can you reason about the question in the OP?
What is the difference between a supertask and
enumerating the algebraic numbers?
On 11/14/2023 5:15 PM, WM wrote:
What is the difference between a supertask and
enumerating the algebraic numbers?
The first is a supertask.
The second is a task.
What is the difference between a supertask and enumerating the algebraic numbers?
On 14.11.2023 23:57, Jim Burns wrote:
On 11/14/2023 5:15 PM, WM wrote:
What is the difference between a supertask and
enumerating the algebraic numbers?
The first is a supertask.
The second is a task.
Why is enumerating the algebraic numbers
only a task?
Why is the game of billiards
more than a task?
On Tuesday, November 14, 2023 at 11:15:05 PM UTC+1, WM wrote:
What is the difference between a supertask and enumerating the algebraic numbers?
1. Spielt bei Supertasks die ZEIT eine Rolle
"a supertask is a countably infinite sequence of operations that occur sequentially within a finite interval of time" (Wikipedia)
2. Geht "enumerating the algebraic numbers" nicht so vonstatten, dass man Zahl für Zahl von Hand durchnummeriert,
On 11/15/2023 5:10 AM, WM wrote:
Why is enumerating the algebraic numbers
only a task?
Why is the game of billiards
more than a task?
The linguistic actions which describe
enumerating the algebraic numbers or
playing your game of billiards have all that,
first, last, last-befores, first-afters.
The indexing actions which are described as
enumerating the algebraic numbers or
playing your game of billiards
_cannot_ have all that.
It is a task to describe either of them.
It is not a task to perform either of them.
On 15.11.2023 19:47, Jim Burns wrote:
On 11/15/2023 5:10 AM, WM wrote:
Why is enumerating the algebraic numbers
only a task?
Why is the game of billiards
more than a task?
The linguistic actions which describe
enumerating the algebraic numbers or
playing your game of billiards have all that,
first, last, last-befores, first-afters.
The indexing actions which are described as
enumerating the algebraic numbers or
playing your game of billiards
_cannot_ have all that.
It is a task to describe either of them.
It is not a task to perform either of them.
So there is no difference.
If all algebraics are enumerated,
then my game is completed too.
"All algebraics are indexed"
is corresponding to my matrix A:
all indices are applied.
On 11/16/2023 3:35 AM, WM wrote:
It is a task to describe either of them.
It is not a task to perform either of them.
So there is no difference.
There is a difference.
On 16.11.2023 13:05, Jim Burns wrote:
On 11/16/2023 3:35 AM, WM wrote:
It is a task to describe either of them.
It is not a task to perform either of them.
So there is no difference.
There is a difference.
All indices are applied to matrix positions.
All indices are applied to roots of polynomials.
What is the difference?
On 11/16/2023 9:23 AM, WM wrote:
There is a difference.
All indices are applied to matrix positions.
All indices are applied to roots of polynomials.
What is the difference?
"All indices are applied to matrix positions"
is not a task.
"All indices are applied to roots of polynomials"
is not a task.
On 16.11.2023 18:03, Jim Burns wrote:
On 11/16/2023 9:23 AM, WM wrote:
It is a task to describe either of them.
It is not a task to perform either of them.
There is a difference.
All indices are applied to matrix positions.
All indices are applied to roots of polynomials.
What is the difference?
"All indices are applied to matrix positions"
is not a task.
"All indices are applied to roots of polynomials"
is not a task.
Call it as you like.
Above you claimed that there is a difference.
What is it?
On 16.11.2023 19:55, Jim Burns wrote:
On 11/16/2023 1:11 PM, WM wrote:
On 16.11.2023 18:03, Jim Burns wrote:
On 11/16/2023 9:23 AM, WM wrote:
It is a task to describe either of them.
It is not a task to perform either of them.
There is a difference.
All indices are applied to matrix positions.
All indices are applied to roots of polynomials.
What is the difference?
"All indices are applied to matrix positions"
is not a task.
"All indices are applied to roots of polynomials"
is not a task.
Call it as you like.
The difference between a task and a supertask
is what you (WM) have been objecting to.
A supertask can disappear Bob.
Spare your nonsense.
Above you claimed that there is a difference.
What is it?
Using finitely-many finite-length claims
to describe a supertask is not itself a supertask.
It is a task.
Maybe.
I asked for the difference between
All indices are applied to matrix positions.
and
All indices are applied to roots of polynomials.
On 11/16/2023 1:11 PM, WM wrote:
On 16.11.2023 18:03, Jim Burns wrote:
On 11/16/2023 9:23 AM, WM wrote:
It is a task to describe either of them.
It is not a task to perform either of them.
There is a difference.
All indices are applied to matrix positions.
All indices are applied to roots of polynomials.
What is the difference?
"All indices are applied to matrix positions"
is not a task.
"All indices are applied to roots of polynomials"
is not a task.
Call it as you like.
The difference between a task and a supertask
is what you (WM) have been objecting to.
A supertask can disappear Bob.
Above you claimed that there is a difference.
What is it?
Using finitely-many finite-length claims
to describe a supertask is not itself a supertask.
It is a task.
On 11/16/2023 2:25 PM, WM wrote:
I asked for the difference between
All indices are applied to matrix positions.
and
All indices are applied to roots of polynomials.
You asked about the difference I claimed above.
I have told you what the difference
I claimed above is. Repeatedly.
It didn't take, apparently.
Indexing the matrix is a supertask.
Describing "indexing the matrix" is a task.
On 16.11.2023 21:01, Jim Burns wrote:
On 11/16/2023 2:25 PM, WM wrote:
I asked for the difference between
All indices are applied to matrix positions.
and
All indices are applied to roots of polynomials.
You asked about the difference I claimed above.
I have told you what the difference
I claimed above is. Repeatedly.
It didn't take, apparently.
Indexing the matrix is a supertask.
Describing "indexing the matrix" is a task.
Yes I have understood that.
Here I apply that knowledge:
Indexing all algebraics is a supertask.
Describing
"indexing all algebraics" is a task.
Correct?
My question however is:
Why can all algebraics be indexed or
be regarded as indexed,
but all matrix positions cannot be indexed or
be regarded as indexed?
but all matrix positions cannot be indexed or
be regarded as indexed?
On 11/16/2023 2:25 PM, WM wrote:
My question however is:
Why can all algebraics be indexed or
be regarded as indexed,
but all matrix positions cannot be indexed or
be regarded as indexed?
You are explicit that you haven't
described indexing each position.
Some positions are dark, you say,
so you can't have done that, you say.
On 17.11.2023 20:13, Jim Burns wrote:
My question however is:
Why can all algebraics be indexed or
be regarded as indexed,
but all matrix positions cannot be indexed or
be regarded as indexed?
You are explicit that you haven't
described indexing each position.
Some positions are dark, you say,
so you can't have done that, you say.
But the main question is this:
is there a difference in the number of indices
applied by me and by Dedekind?
On 11/17/2023 2:25 PM, WM wrote:
But the main question is this:
is there a difference in the number of indices
applied by me and by Dedekind?
No,
that doesn't address the question asked.
The same indices can index all the positions
(of B, for example.)
and, applied differently,
NOT index all the positions.
On 17.11.2023 20:56, Jim Burns wrote:
On 11/17/2023 2:25 PM, WM wrote:
My question however is:
Why can all algebraics be indexed or
be regarded as indexed,
but all matrix positions cannot be indexed or
be regarded as indexed?
But the main question is this:
is there a difference in the number of indices
applied by me and by Dedekind?
No,
that doesn't address the question asked.
But it addresses the question I asked.
The same indices can index all the positions
(of B, for example.)
and, applied differently,
NOT index all the positions.
In case of enumerating the fractions
I apply the indices just like Cantor does
and show that there remain
not indexed positions surrounding B.
Most of your symbols are not readable.
Please constrict the use.
My use of symbols is brilliant and delightful.
On 11/17/2023 4:37 PM, WM wrote:
In case of enumerating the fractions
I apply the indices just like Cantor does
and show that there remain
not indexed positions surrounding B.
No,
you don't show that.
You claim it.
On 18.11.2023 00:46, Jim Burns wrote:
On 11/17/2023 4:37 PM, WM wrote:
In case of enumerating the fractions
I apply the indices just like Cantor does
and show that there remain
not indexed positions surrounding B.
No,
you don't show that.
You claim it.
I use his formula
k = (m + n - 1)(m + n - 2)/2 + m
and index the positions
where the fractions
1/1, 1/2, 2/1, 1/3, [...]
initially sit.
On 11/18/2023 3:35 AM, WM wrote:
On 18.11.2023 00:46, Jim Burns wrote:
On 11/17/2023 4:37 PM, WM wrote:
In case of enumerating the fractions
I apply the indices just like Cantor does
and show that there remain
not indexed positions surrounding B.
No,
you don't show that.
You claim it.
I use his formula
k = (m + n - 1)(m + n - 2)/2 + m
and index the positions
You and Cantor don't literally index those positions.
All the indexing would be a supertask.
You and Cantor describe the indexes of those positions.
All the describing is a task.
where the fractions
1/1, 1/2, 2/1, 1/3, [...]
initially sit.
And you have described, for each fraction m/n,
its index k
Each of infinitely-many of them.
Complete.
Where have you shown what you say you've shown?
for each integer m ≥ 1
and each integer n ≥ 1 there are
unique integer sₘₙ = m+n ≥ 2
Each fraction m/n has
one index kₘₙ
On 18.11.2023 00:46, Jim Burns wrote:
On 11/17/2023 4:37 PM, WM wrote:
Your use of symbols may satisfy you
but they cannot be read and
therefore are no means of communication.
On 18.11.2023 19:45, Jim Burns wrote:
On 11/18/2023 3:35 AM, WM wrote:
On 18.11.2023 00:46, Jim Burns wrote:
On 11/17/2023 4:37 PM, WM wrote:
In case of enumerating the fractions
I apply the indices just like Cantor does
and show that there remain
not indexed positions surrounding B.
No,
you don't show that.
You claim it.
I use his formula
k = (m + n - 1)(m + n - 2)/2 + m
and index the positions
You and Cantor don't literally index those positions.
All the indexing would be a supertask.
Of course only the first indexing can be done.
Therefore Cantor's claim
is pure nonsense.
You and Cantor describe the indexes of
those positions.
Anyhow we do the same.
If Cantor fails to issue all indices,
then I fail too.
If he succeeds, then I succeed too.
And I show that fractions occupy
not indexed positions of the matrix.
Where have you shown what you say you've shown?
Here: 1, 1/2, 1/3, 1/4, ...
2, 2/2, 2/3, 2/4, ...
3, 3/2, 3/3, 3/4, ...
4, 4/2, 4/3, 4/4, ...
5, 5/2, 5/3, 5/4, ...
Push the natnumbers of the first column
(without queue) into the field of fractions
and store the hit fraction always there
where the natnumber has come from.
Try to push the natnumbers such that
all matrix positions are occupied by them.
That is best done by creating a pattern like
1, 2, 4, ...
3, 5, 8, ...
6, 9, 13, ...
... ,
According to this simple rule
it is impossible, in eternity,
to remove a fraction from the matrix or
to attach a natnumber to a fraction.
No one else reads Cantor's claim as a claim
to have literally performed a supertask.
It is nonsense to claim
to have performed a supertask,
unless you are Chuck Norris.
Georg Cantor is not Chuck Norris.
On 11/18/2023 2:49 PM, WM wrote:
On 18.11.2023 19:45, Jim Burns wrote:
On 11/18/2023 3:35 AM, WM wrote:
On 18.11.2023 00:46, Jim Burns wrote:
On 11/17/2023 4:37 PM, WM wrote:
In case of enumerating the fractions
I apply the indices just like Cantor does
and show that there remain
not indexed positions surrounding B.
No,
you don't show that.
You claim it.
I use his formula
k = (m + n - 1)(m + n - 2)/2 + m
and index the positions
You and Cantor don't literally index those positions.
All the indexing would be a supertask.
Of course only the first indexing can be done.
All the indexing is described.
Therefore Cantor's claim is pure nonsense.
No one else reads Cantor's claim as a claim
to have literally performed a supertask.
It is nonsense to claim
to have performed a supertask,
You and Cantor describe the indexes of
those positions.
Anyhow we do the same.
If Cantor fails to issue all indices,
then I fail too.
If he succeeds, then I succeed too.
You both fail to perform supertasks.
That is not unexpected.
And I show that fractions occupy
not indexed positions of the matrix.
You do not show that.
Non-literally, you push fractions and
indices around and then _declare_ that
fractions are still in the matrix.
Your description of pushing fractions and
indices around does not forget any fractions,
and does not assign any index twice.
Your _declaration_ that
fractions are still in the matrix
is wrong.
Where have you shown what you say you've shown?
Here: 1, 1/2, 1/3, 1/4, ...
2, 2/2, 2/3, 2/4, ...
3, 3/2, 3/3, 3/4, ...
4, 4/2, 4/3, 4/4, ...
5, 5/2, 5/3, 5/4, ...
Push the natnumbers of the first column
(without queue) into the field of fractions
and store the hit fraction always there
where the natnumber has come from.
Try to push the natnumbers such that
all matrix positions are occupied by them.
Push kᵢⱼ to ⟨i,j⟩ such that
kᵢⱼ = i+(i+j-1)(i+j-2)/2
Push them all.
Pushing them all is a supertask.
If any position is not indexed,
all have not been pushed.
Equivalently,
if all have been pushed,
each position has been indexed.
That is best done by creating a pattern like
1, 2, 4, ...
3, 5, 8, ...
6, 9, 13, ...
... ,
According to this simple rule
it is impossible, in eternity,
to remove a fraction from the matrix or
to attach a natnumber to a fraction.
Pushing them all is a supertask.
Describing all that and
finitely augmenting all that not-first-false-ly
is a task.
On Saturday, November 18, 2023 at 11:53:05 PM UTC+1, Jim Burns wrote:
No one else reads Cantor's claim as a claim
to have literally performed a supertask.
It is nonsense to claim
to have performed a supertask,
unless you are Chuck Norris.
Georg Cantor is not Chuck Norris.
Completely agree with you, for once.
On 11/18/2023 3:35 AM, WM wrote:
On 18.11.2023 00:46, Jim Burns wrote:
On 11/17/2023 4:37 PM, WM wrote:
Your use of symbols may satisfy you
but they cannot be read and
therefore are no means of communication.
You (WM) have so far not asked for
the meaning of any of the symbols I've used.
is much harder to read, even in English, than
| ∀n ∈ ℕ\{0,1,2}: ¬∃x,y,z ∈ ℕ\{0}: xⁿ+yⁿ = zⁿ
But it may take symbols crafted for a particular
context to fit such large amounts of content into
such small spaces.
I don't wish to stop (for example) using ℕˡᵘᵇ
to refer to the set of numbers I am talking about,
because
I am talking about the particular set for which
∀S: ∀⟨1,…,n⟩ ⊆ S ⟹ ∀⟨1,…,n⟩ ⊆ ℕˡᵘᵇ ⊆ S
a set which does not hold darkᵂᴹ numbers and
also a set which satisfies the claims which
you say require darkᵂᴹ numbers.
On 18.11.2023 23:53, Jim Burns wrote:
On 11/18/2023 2:49 PM, WM wrote:
On 18.11.2023 19:45, Jim Burns wrote:
On 11/18/2023 3:35 AM, WM wrote:
On 18.11.2023 00:46, Jim Burns wrote:
On 11/17/2023 4:37 PM, WM wrote:
In case of enumerating the fractions
I apply the indices just like Cantor does
and show that there remain
not indexed positions surrounding B.
No,
you don't show that.
You claim it.
I use his formula
k = (m + n - 1)(m + n - 2)/2 + m
and index the positions
You and Cantor don't literally index
those positions.
All the indexing would be a supertask.
Of course only the first indexing can be done.
All the indexing is described.
Since
in my application of Cantor's algorithm
never any fraction is siiting at an indexed place
but all fractions remain in the matrix,
not all indexing can be described.
Your description of pushing fractions and
indices around does not forget any fractions,
and does not assign any index twice.
Your _declaration_ that
fractions are still in the matrix
is wrong.
If you deny mathematics and logic completely,
then it is impossible to prove anything
On 11/19/2023 6:05 AM, WM wrote:
Since
in my application of Cantor's algorithm
never any fraction is sitting at an indexed place
but all fractions remain in the matrix,
not all indexing can be described.
I describe all and only these swaps
On 20.11.2023 01:29, Jim Burns wrote:
On 11/19/2023 6:05 AM, WM wrote:
Since
in my application of Cantor's algorithm
never any fraction is sitting at an indexed place
but all fractions remain in the matrix,
not all indexing can be described.
I describe all and only these swaps
Whatever you describe:
Fact is that every matrix A(i) with A(0) =
1, 1/2, 1/3, 1/4, ...
2, 2/2, 2/3, 2/4, ...
3, 3/2, 3/3, 3/4, ...
4, 4/2, 4/3, 4/4, ...
5, 5/2, 5/3, 5/4, ...
...
has the same extension.
In all matrices A(i) the fractions remain on
positions without index, by definition.
Therefore
the matrices B(i) of indices in their final places
cannot be the complete matrices.
On Monday, November 20, 2023 at 8:05:43 PM UTC+1, Jim Burns wrote: [...]
I'd promised you a _mathematical_ approach which might be adopted as a "framework" for (some of) your considerations.
It would even allow to _deduce_ "by Bob".
But I'm still pondering how to ...
On 11/20/2023 6:47 AM, WM wrote:
Fact is that every matrix A(i) with A(0) =
1, 1/2, 1/3, 1/4, ...
2, 2/2, 2/3, 2/4, ...
3, 3/2, 3/3, 3/4, ...
4, 4/2, 4/3, 4/4, ...
5, 5/2, 5/3, 5/4, ...
...
has the same extension.
extension == set of positions
In all matrices A(i) the fractions remain on positions without index,
by definition.
I describe kᵢⱼ for each ⟨i,j⟩ in ℕ×ℕ such that
no kᵢⱼ is in two positions.
I don't merely _claim_ that claim.
I start with a description
Therefore
the matrices B(i) of indices in their final places
cannot be the complete matrices.
No B(i) is the result of all the swaps described.
Each swap described is accessible and
followed in the swaps described.
All the swaps is not a task.
On 20.11.2023 20:05, Jim Burns wrote:
On 11/20/2023 6:47 AM, WM wrote:
Since
in my application of Cantor's algorithm
never any fraction is sitting at an indexed place
but all fractions remain in the matrix,
not all indexing can be described.
In all matrices A(i)
the fractions remain on positions without index,
by definition.
I describe kᵢⱼ for each ⟨i,j⟩ in ℕ×ℕ such that
no kᵢⱼ is in two positions.
I don't merely _claim_ that claim.
I start with a description
So do I. See the OP.
Therefore
the matrices B(i) of indices in their final places
cannot be the complete matrices.
No B(i) is the result of all the swaps described.
Each swap described is accessible and
followed in the swaps described.
No term of Cantor's sequence describes
a complete bijection.
All the swaps is not a task.
But *all* terms of the sequence 1/1, 1/2, [...]
are required for a bijection.
...less positions...
On 11/21/2023 5:55 AM, WM wrote:
I start with a description
So do I. See the OP.
and I augment it with not-first-false claims
-- which must be true as well as not-first-false
because
they are in a finite sequence of claims in which,
because it's finite,
if any claim is false, some claim is first-false,
and no claim in that sequence is first-false.
I claim
in a finite only-not-first-false claim-sequence
that
kᵢⱼ exists for each ⟨i,j⟩ in ℕ×ℕ such that
no kᵢⱼ is in two positions
Unless I'm wrong about the nature of
the claim-sequence, my claim is true.
followed in the swaps described.
No term of Cantor's sequence describes
a complete bijection.
The terms are not the description I give.
Indexing *all* the terms is a supertask.
You don't do that.
No one ever does that
Describing indexing *all* the terms is a task.
We do that.
We learn about *all* the terms by
augmenting the description with
not-first-only claims.
We do that, too.
...less positions...
| How many positions? Fewer positions.
or
| How much position? Less position.
"fewer /count noun/"
count nouns (zählbares Substantiv) have plurals
"less /mass noun/"
mass nouns (Stoffname) don't have plurals
The same noun may appear in either
linguistic role, which doesn't guarantee
two ways of making sense.
My game tells us
On Tuesday, November 21, 2023 at 8:19:25 PM UTC+1, WM wrote:why sci dot logic is dead.
My game tells us
nothing, you psychotic asshole full of shit.
On Wednesday, November 22, 2023 at 8:38:27 PM UTC-5, Fritz Feldhase wrote:
On Tuesday, November 21, 2023 at 8:19:25 PM UTC+1, WM wrote:why sci dot logic is dead.
My game tells us
On 11/27/2023 6:52 AM, WM wrote:
Jim Burns schrieb am Montag,
X contains only those natnumbers which,
when added to X,
leave |ℕ \ X| = ℵo.
X contains only those natnumbers which,
already being in X,
cannot be added to X
X is a set function.X is a collection.
You, upthread in sci.math
Collect
all natnumbers n with the above property
into a collection X
Collecting those is a supertask, not a task.
It can be increased.
A set increased from X
is not X
But never |ℕ \ X| < ℵo will be accomplishedWe who are not Chuck Norris can't do supertasks.
by adding individually definable numbers.
On Monday, November 27, 2023 at 12:53:01 PM UTC+1, WM wrote:everything is a _set_.]
X is a set function.Nope. You defined X as a "collection" (set, class). [Hint: In ZF(C)
It can be increased.
Nope. Sets ("collections") do neither grow nor shrink.
Jim Burns schrieb am Montag,Newsgroups: sci.logic
27. November 2023 um 19:05:05 UTC+1:
You, upthread in sci.math
Collect
all natnumbers n with the above property
into a collection X
Collecting those is a supertask, not a task.
No, there are only less than ℵo.
But never |ℕ \ X| < ℵo will be accomplished
by adding individually definable numbers.
We who are not Chuck Norris can't do supertasks.
Collect as many as you can.
Never |ℕ \ X| < ℵo will be accomplished.
But |ℕ \ ℕ| = ℵo ==>
ℕ contains dark numbers.
X is a set function.
X is a function. The arguments has been dropped.
On 11/27/2023 2:36 PM, WM wrote:
Each 1×1 1.ended is the same "size" ℵ₀
Collect as many as you can.
Fewer than 10^(10^(10^(10^(10))))
Infinitely fewer than we can reason about.
Define X
Fritz Feldhase schrieb am Dienstag, 28. November 2023 um 01:57:37 UTC+1:
Define X
On Tuesday, November 28, 2023 at 10:41:32 AM UTC+1, WM wrote:
Fritz Feldhase schrieb am Dienstag, 28. November 2023 um 01:57:37 UTC+1:
Define X
Ich hatte geschrieben:
| 1. Define the notion /set function/
| 2. Define X
| 3. Show that X is a set function (based on the definitions 1 and 2)
On 28.11.2023 15:22, Fritz Feldhase wrote:
| 1. Define the notion /set function/
A set function is a sequence of sets S(n) depending on an argument n. Example: The sequence of FISONs (F(n)) with F(n) = {1, 2, 3, ..., n}.
| 2. Define X
X is the sequence and union of FISONs
| 3. Show that X is a set function (based on the definitions 1 and 2)
Obvious. See above.
On 27.11.2023 22:11, Jim Burns wrote:
Each 1×1 1.ended is the same "size" ℵ₀
ℵ₀ is not a size
but a shorthand for actual infinity.
ℵ₀ is not a size
but a shorthand for actual infinity.
|ℕ| = ℵ₀ as well
as |ℚ| = 2|ℕ|^2 + 1 = ℵ₀ .
Collect as many as you can.
Fewer than 10^(10^(10^(10^(10))))
Infinitely fewer than we can reason about.
If you can give a number like above,
then all smaller numbers are
automatically defined too.
All last numbers of a FISON that
you can reason about and
all their predecessors
all their predecessors belong to X.
You cannot reason about numbers as individuals
which are in the difference |ℕ \ X|,
can you?
You cannot reason about numbers as individuals
which are in the difference |ℕ \ X|,
can you?
Fritz Feldhase schrieb am Dienstag, 28. November 2023 um 01:57:37 UTC+1:
Define X
If you can identify a natnumber, then this number and all smaller
numbers are automatically elements of X. (All last numbers of FISONs
that you can reason about and all their predecessors belong to X.) You
cannot reason about natnumbers as individuals which are in the
difference |ℕ \ X| = ℵo. As soon as you identify a natnumber there, it belongs to X. Nevertheless the difference remains actually infinite: ℵo natnumbers. Therefore they are dark.
Regards, WM
On 11/28/23 4:41 AM, WM wrote:is there a less than relation to your set?
Fritz Feldhase schrieb am Dienstag, 28. November 2023 um 01:57:37 UTC+1:
Define X
If you can identify a natnumber, then this number and all smaller
numbers are automatically elements of X. (All last numbers of FISONs
that you can reason about and all their predecessors belong to X.) You cannot reason about natnumbers as individuals which are in the
difference |ℕ \ X| = ℵo. As soon as you identify a natnumber there, it belongs to X. Nevertheless the difference remains actually infinite: ℵo natnumbers. Therefore they are dark.
Regards, WMExcept that this logic is flawed. What is the first number in that difference, it is the highest number that we can reason about/identify + 1.
Since we CAN reason about that, since we just did, then this number
needs to be in X and not be a "dark" number.
Your logic INCORRECTLY presumes that there IS a finite number that is
the highest we can identify or reason about. There is no actual basis
for assuming that such a number must exist, and the assumption is shown
to make the system inconsistant (as we have a "dark" number that is identified, so it isn't actually dark).
On Tuesday, November 28, 2023 at 9:04:06 PM UTC-5, Richard Damon wrote:your visible set is ordered.
On 11/28/23 4:41 AM, WM wrote:
Fritz Feldhase schrieb am Dienstag, 28. November 2023 um 01:57:37 UTC+1:
Define X
If you can identify a natnumber, then this number and all smaller numbers are automatically elements of X. (All last numbers of FISONs that you can reason about and all their predecessors belong to X.) You cannot reason about natnumbers as individuals which are in the difference |ℕ \ X| = ℵo. As soon as you identify a natnumber there, it
belongs to X. Nevertheless the difference remains actually infinite: ℵo
natnumbers. Therefore they are dark.
Regards, WMExcept that this logic is flawed. What is the first number in that difference, it is the highest number that we can reason about/identify + 1.
Since we CAN reason about that, since we just did, then this number
needs to be in X and not be a "dark" number.
Your logic INCORRECTLY presumes that there IS a finite number that isis there a less than relation to your set?
the highest we can identify or reason about. There is no actual basis
for assuming that such a number must exist, and the assumption is shown
to make the system inconsistant (as we have a "dark" number that is identified, so it isn't actually dark).
that would necessitate a finite structure to perform comparison.
without lessthan the set may contain aleph null natnumbers.
daniel2383
On Tuesday, November 28, 2023 at 10:44:26 PM UTC-5, Daniel Pehoushek wrote:two members of the larger set are not comparable with less than.
On Tuesday, November 28, 2023 at 9:04:06 PM UTC-5, Richard Damon wrote:
On 11/28/23 4:41 AM, WM wrote:
Fritz Feldhase schrieb am Dienstag, 28. November 2023 um 01:57:37 UTC+1:
Define X
If you can identify a natnumber, then this number and all smaller numbers are automatically elements of X. (All last numbers of FISONs that you can reason about and all their predecessors belong to X.) You cannot reason about natnumbers as individuals which are in the difference |ℕ \ X| = ℵo. As soon as you identify a natnumber there, it
belongs to X. Nevertheless the difference remains actually infinite: ℵo
natnumbers. Therefore they are dark.
Regards, WMExcept that this logic is flawed. What is the first number in that difference, it is the highest number that we can reason about/identify + 1.
Since we CAN reason about that, since we just did, then this number needs to be in X and not be a "dark" number.
your visible set is ordered.Your logic INCORRECTLY presumes that there IS a finite number that is the highest we can identify or reason about. There is no actual basis for assuming that such a number must exist, and the assumption is shown to make the system inconsistant (as we have a "dark" number that is identified, so it isn't actually dark).is there a less than relation to your set?
that would necessitate a finite structure to perform comparison.
without lessthan the set may contain aleph null natnumbers.
daniel2383
the larger set is not.
avoid negation and prosper in truth
daniel2383
On Tuesday, November 28, 2023 at 10:44:26 PM UTC-5, Daniel Pehoushek
is there a less than relation to your set?
that would necessitate a finite structure to perform comparison.your visible set is ordered.
without lessthan the set may contain aleph null natnumbers.
daniel2383
the larger set is not.
On 11/28/23 4:41 AM, WM wrote:
Fritz Feldhase schrieb am Dienstag, 28. November 2023 um 01:57:37 UTC+1:
Define X
If you can identify a natnumber, then this number and all smaller
numbers are automatically elements of X. (All last numbers of FISONs
that you can reason about and all their predecessors belong to X.) You
cannot reason about natnumbers as individuals which are in the
difference |ℕ \ X| = ℵo. As soon as you identify a natnumber there, it >> belongs to X. Nevertheless the difference remains actually infinite:
ℵo natnumbers. Therefore they are dark.
What is the first number in that
difference, it is the highest number that we can reason about/identify + 1.
Since we CAN reason about that, since we just did, then this number
needs to be in X and not be a "dark" number.
Your logic INCORRECTLY presumes that there IS a finite number that is
the highest we can identify or reason about.
two members of the larger set are not comparable with less than.
Daniel Pehoushek schrieb am Mittwoch, 29. November 2023 um 05:09:26 UTC+1:
is there a less than relation to your set?
Not to the dark numbers.
your visible set is ordered. the larger set is not.
So it is.
On Wednesday, November 29, 2023 at 12:56:14 PM UTC+1, WM wrote:i am saying < is an algorithm on two finite structures
Daniel Pehoushek schrieb am Mittwoch, 29. November 2023 um 05:09:26 UTC+1:
is there a less than relation to your set?
Not to the dark numbers.Which implies that your "dark numbers" aren't natural numbers.
Hint: Any 'two' natural numbers n, m are /comparable/ (with respect to <=).
In other words, for all n, m e IN: n <= m or m <= n.
your visible set is ordered. the larger set is not.
So it is.Then "the larger set" is not IN.
Hint: IN is a totally ordered set (with respect to <=).
See: https://en.wikipedia.org/wiki/Total_order
Actually, (due to Cantor) it is a well-ordered set, you know.
See: https://en.wikipedia.org/wiki/Well-order
i am saying
On 29.11.2023 03:04, Richard Damon wrote:
On 11/28/23 4:41 AM, WM wrote:
Fritz Feldhase schrieb am Dienstag, 28. November 2023 um 01:57:37 UTC+1: >>>
Define X
If you can identify a natnumber, then this number and all smaller
numbers are automatically elements of X. (All last numbers of FISONs
that you can reason about and all their predecessors belong to X.)
You cannot reason about natnumbers as individuals which are in the
difference |ℕ \ X| = ℵo. As soon as you identify a natnumber there,
it belongs to X. Nevertheless the difference remains actually
infinite: ℵo natnumbers. Therefore they are dark.
What is the first number in that difference, it is the highest number
that we can reason about/identify + 1.
Dark numbers cannot be used as individuals. The visible nunbers are potentially infinite. There is no last one. So there is no first dark one.
Since we CAN reason about that, since we just did, then this number
needs to be in X and not be a "dark" number.
No. Visible numbers n have finite initial segments {1, 2, 3, ..., n}.
Your logic INCORRECTLY presumes that there IS a finite number that is
the highest we can identify or reason about.
No. With every vivible number n also n^n^n is visible. Pot. Inf.
Regards, WM
On Wednesday, November 29, 2023 at 7:30:19 AM UTC-5, Fritz Feldhase wrote:
On Wednesday, November 29, 2023 at 12:56:14 PM UTC+1, WM wrote:
Daniel Pehoushek schrieb am Mittwoch, 29. November 2023 um 05:09:26 UTC+1: >>>>Which implies that your "dark numbers" aren't natural numbers.
is there a less than relation to your set?Not to the dark numbers.
Hint: Any 'two' natural numbers n, m are /comparable/ (with respect to <=). >>
In other words, for all n, m e IN: n <= m or m <= n.
Then "the larger set" is not IN.your visible set is ordered. the larger set is not.So it is.
On 11/29/23 6:50 AM, WM wrote:
Visible numbers n have finite initial segments {1, 2, 3, ..., n}.
Your logic INCORRECTLY presumes that there IS a finite number that is
the highest we can identify or reason about.
No. With every visible number n also n^n^n is visible. Pot. Inf.
So, all you have done is shown that your "dark numbers" aren't
"numbers", and thus don't exist in the initial set that they are a
remainder of, and thus are just an empty set.
On 30.11.2023 04:42, Richard Damon wrote:
On 11/29/23 6:50 AM, WM wrote:
Visible numbers n have finite initial segments {1, 2, 3, ..., n}.
Your logic INCORRECTLY presumes that there IS a finite number that
is the highest we can identify or reason about.
No. With every visible number n also n^n^n is visible. Pot. Inf.
So, all you have done is shown that your "dark numbers" aren't
"numbers", and thus don't exist in the initial set that they are a
remainder of, and thus are just an empty set.
Wrong.
ℕ_vis contains numbers which have ℵo successors. ℕ contains all numbers,
also these successors. The successors cannot be well-ordered because
they follow upon every number that belongs to a well-order. Therefore
there is definitely a difference.
Regards, WM
= x, and for any member of ℕ such a number exists.
So, all you have done is shown that your "dark numbers" aren't
"numbers", and thus don't exist in the initial set that they are a
remainder of, and thus are just an empty set.
On 30.11.2023 04:42, Richard Damon wrote:
So, all you have done is shown that
your "dark numbers" aren't "numbers",
and thus
don't exist in the initial set that
they are a remainder of,
and thus
are just an empty set.
Here is a proof that
you will easier understand:
Between
every definable unit fraction 1/n and 0,
there are ℵo smaller unit fractions.
You cannot reduce this amount to less than ℵo.
You cannot distinguish ℵo of them.
But they must exist in the interval (0,1].
So they are not an empty set, but existing.
On 30.11.2023 04:42, Richard Damon wrote:
So, all you have done is shown that your "dark numbers" aren't
"numbers", and thus don't exist in the initial set that they are a
remainder of, and thus are just an empty set.
Here is a proof that you will easier understand:
Between every definable unit fraction 1/n and 0, there are ℵo smaller
unit fractions. You cannot reduce this amount to less than ℵo. You
cannot distinguish ℵo of them. But they must exist in the interval (0,
1]. So they are not an empty set, but existing.
Regards, WM
On Friday, December 1, 2023 at 1:30:44 PM UTC+1, WM wrote:
Between every [...] unit fraction 1/n and 0, there are ℵo smaller unit fractions.
Indeed!
You cannot reduce this amount to less than ℵo.
It's just a fakt. Period.
You cannot distinguish ℵo of them.
??? Actually, the ARE already "distinguished".
No matter if *I* can "distinguish ℵo of them" or not.
For all definable 1/n we get:But they must exist in the interval (0, 1].
Right. They do. :-)
Hint: An e IN: 0 < 1/n <= 1.
So they are not an empty set, but existing.
Sure.
{1/n : n e IN} =/= { }.
Actually,
card {1/n : n e IN} = ℵo.
ℵo unit fractions lie at the left-hand side of (1/n, 1].
ℵo unit fractions lie at the left-hand side of (0, 1].
On 12/1/23 7:30 AM, WM wrote:
On 30.11.2023 04:42, Richard Damon wrote:
So, all you have done is shown that your "dark numbers" aren't
"numbers", and thus don't exist in the initial set that they are a
remainder of, and thus are just an empty set.
Here is a proof that you will easier understand:
Between every definable unit fraction 1/n and 0, there are ℵo smaller
unit fractions. You cannot reduce this amount to less than ℵo. You
cannot distinguish ℵo of them. But they must exist in the interval (0,
1]. So they are not an empty set, but existing.
Why do you say you can not "distinguish" them,
they are merely the unit
fractions 1/x for all Natural Numbers x greater than n (all ℵo of them).
There are ℵo Natural numbers, and removing n of them, still leaves you
with ℵo of them (that's how mathematics of trans-finite numbers work).
There is no need for some "dark" numbers to explain this, they are just
the orinary natural numbers.
What "Unit Fraction" exists in that interval that isn't one of those 1/x's?
Yes, no matter how high you choose for n, you still have ℵo numbers
abort it, but that is just the nature of infinite sets, which I guess is above your ability to understand.
ℵo unit fractions lie at the left-hand side of (eps, 1].
ℵo unit fractions lie at the left-hand side of (0, 1].
On 02.12.2023 00:41, Richard Damon wrote:
On 12/1/23 7:30 AM, WM wrote:
On 30.11.2023 04:42, Richard Damon wrote:
So, all you have done is shown that your "dark numbers" aren't
"numbers", and thus don't exist in the initial set that they are a
remainder of, and thus are just an empty set.
Here is a proof that you will easier understand:
Between every definable unit fraction 1/n and 0, there are ℵo smaller
unit fractions. You cannot reduce this amount to less than ℵo. You
cannot distinguish ℵo of them. But they must exist in the interval
(0, 1]. So they are not an empty set, but existing.
Why do you say you can not "distinguish" them,
Because it is fact. You cannot reduce this amount to less than ℵo.
they are merely the unit fractions 1/x for all Natural Numbers x
greater than n (all ℵo of them).
I do not deny that they exist. But I denay that they can be used as individuals. They can only be used collectively. I call them dark.
There are ℵo Natural numbers, and removing n of them, still leaves you
with ℵo of them (that's how mathematics of trans-finite numbers work).
You cannot reduce this amount to less than ℵo:
∀n ∈ ℕ_def: |ℕ \ {1, 2, 3, ..., n}| = ℵo.
But you can use them collectively:
|ℕ \ {1, 2, 3, ...}| = 0.
They are dark.
There is no need for some "dark" numbers to explain this, they are
just the orinary natural numbers.
But it is fact that they cannot be used as individuals. Therefore all of Cantor's "bijections" are wrong. He assumes that all natnumbers can be
used as individuals.
What "Unit Fraction" exists in that interval that isn't one of those
1/x's?
Yes, no matter how high you choose for n, you still have ℵo numbers
abort it, but that is just the nature of infinite sets, which I guess
is above your ability to understand.
If you are better, then try to explain this:
For all definable eps > 0 we get:
∀x ∈ (eps, 1] are larger than ℵo unit fractions.
ℵo unit fractions lie at the left-hand side of (eps, 1].
For all x > 0 we get:
∀x ∈ (0, 1] are larger than ℵo unit fractions.
ℵo unit fractions lie at the left-hand side of (0, 1].
Why this difference?
Gruß, WM
On 12/2/23 5:43 AM, WM wrote:
Why do you say you can not "distinguish" them,
Because it is fact. You cannot reduce this amount to less than ℵo.
And why you need to?
I do not deny that they exist. But I deny that they can be used as
individuals. They can only be used collectively. I call them dark.
Which number x greater than your first chosen n can't be used as an individual?
There are ℵo Natural numbers, and removing n of them, still leaves
you with ℵo of them (that's how mathematics of trans-finite numbers
work).
You cannot reduce this amount to less than ℵo:
∀n ∈ ℕ_def: |ℕ \ {1, 2, 3, ..., n}| = ℵo.
But you can use them collectively:
|ℕ \ {1, 2, 3, ...}| = 0.
They are dark.
Again, why can't you use any of those individually.
There is no need for some "dark" numbers to explain this, they are
just the orinary natural numbers.
But it is fact that they cannot be used as individuals. Therefore all
of Cantor's "bijections" are wrong. He assumes that all natnumbers can
be used as individuals.
And which ones can't be?
If you are better, then try to explain this:
For all definable eps > 0 we get:
∀x ∈ (eps, 1] are larger than ℵo unit fractions.
ℵo unit fractions lie at the left-hand side of (eps, 1].
For all x > 0 we get:
∀x ∈ (0, 1] are larger than ℵo unit fractions.
ℵo unit fractions lie at the left-hand side of (0, 1].
Why this difference?
This just comes down to the fact that there isn't a "last" natural
number or a "first" unit fraction.
You just seem to think becuase we can't "name" the highest natural
number (because there isn't one) at some point they become "dark".
It seems your mind just can't handle infinity. Which is actually a
fairly common problem.
On 12/3/23 6:38 AM, WM wrote:
On 02.12.2023 16:45, Richard Damon wrote:
you still have ℵo
natural numbers you can use,
Yes, but it explains the problem vividly.And why is that a problem? That is just a simple fact that happens when
you move into infinites.
ℵo unit fractions lie at the left-hand side of (0, 1].stupid
No, the problem is that most mathematicians are too naive or too
theto recognize that all points at the left-hand side of x are also at
left-hand side of the interval [x, oo). Hence the claim that ℵo unit fractions lie at the left-hand side of every x > 0 would place ℵo unit fractions below zero, which is nonsense.Why do you claim that. That is based on the FALSE assumption that their
IS a "first unit fraction", which implies that there is a "Highest
Natural Number".
Your additional unit fractions lie in the interval of
(0, x), which has room for an infinite number of unit fractions as long
as x > 0.
Your unit fractions are "dense" near 0, which means there is
no "first" unit fraction.
On 02.12.2023 16:45, Richard Damon wrote:
On 12/2/23 5:43 AM, WM wrote:
Why do you say you can not "distinguish" them,
Because it is fact. You cannot reduce this amount to less than ℵo.
And why you need to?
To set up a bijection or enumeration.
I do not deny that they exist. But I deny that they can be used as
individuals. They can only be used collectively. I call them dark.
Which number x greater than your first chosen n can't be used as an
individual?
Every such number can be used as an individual. With n also n^n^n can be used. But almost all cannot.
There are ℵo Natural numbers, and removing n of them, still leaves
you with ℵo of them (that's how mathematics of trans-finite numbers
work).
You cannot reduce this amount to less than ℵo:
∀n ∈ ℕ_def: |ℕ \ {1, 2, 3, ..., n}| = ℵo.
But you can use them collectively:
|ℕ \ {1, 2, 3, ...}| = 0.
They are dark.
Again, why can't you use any of those individually.
Because after every attempt almost all remain. That's how mathematics of transfinite numbers works. But set theorists claim that all can be used
for enumerating.
There is no need for some "dark" numbers to explain this, they are
just the orinary natural numbers.
But it is fact that they cannot be used as individuals. Therefore all
of Cantor's "bijections" are wrong. He assumes that all natnumbers
can be used as individuals.
And which ones can't be?
After every attempt almost all remain. They cannot be used.
If you are better, then try to explain this:
For all definable eps > 0 we get:
∀x ∈ (eps, 1] are larger than ℵo unit fractions.
ℵo unit fractions lie at the left-hand side of (eps, 1].
For all x > 0 we get:
∀x ∈ (0, 1] are larger than ℵo unit fractions.
ℵo unit fractions lie at the left-hand side of (0, 1].
Why this difference?
This just comes down to the fact that there isn't a "last" natural
number or a "first" unit fraction.
Yes, but it explains the problem vividly.
You just seem to think becuase we can't "name" the highest natural
number (because there isn't one) at some point they become "dark".
I know that below 0, there are no unit fractions. With increasing x > 0
their number increases. But all have finite distances between each
other: ∀n ∈ ℕ: 1/n - 1/(n+1) = d_n > 0. Therefore there must be one the first.
It seems your mind just can't handle infinity. Which is actually a
fairly common problem.
No, the problem is that most mathematicians are too naive or too stupid
to recognize that all points at the left-hand side of x are also at the left-hand side of the interval [x, oo). Hence the claim that ℵo unit fractions lie at the left-hand side of every x > 0 would place ℵo unit fractions below zero, which is nonsense.
Regards, WM
Richard Damon schrieb am Sonntag, 3. Dezember 2023 um 13:33:22 UTC+1:
On 12/3/23 6:38 AM, WM wrote:
On 02.12.2023 16:45, Richard Damon wrote:
you still have ℵo
natural numbers you can use,
You cannot use them, because every attempt you make will leave them unused.
Yes, but it explains the problem vividly.And why is that a problem? That is just a simple fact that happens when you move into infinites.
There is no way to "move to infinites". Further mathematics remains
valid in every case: ∀x ∈ (0, 1] are larger than ℵo unit fractions.
ℵo unit fractions lie at the left-hand side of (0, 1].
stupidNo, the problem is that most mathematicians are too naive or too
theto recognize that all points at the left-hand side of x are also at
left-hand side of the interval [x, oo). Hence the claim that ℵo unit fractions lie at the left-hand side of every x > 0 would place ℵo unit fractions below zero, which is nonsense.Why do you claim that. That is based on the FALSE assumption that their
IS a "first unit fraction", which implies that there is a "Highest
Natural Number".
Because it is mathematical fact. What is at the left-hand side of x is
also at the left-hand side of the interval [x, oo).
To deny this is non-mathematical nonsense.
Your additional unit fractions lie in the interval of
(0, x), which has room for an infinite number of unit fractions as long
as x > 0.
No.
Your unit fractions are "dense" near 0, which means there is
no "first" unit fraction.
They are not dense enough to avoid the finite intervals between them.
They are linearly ordered which means that there is a first one after
zero. The only alternative would be many first ones simultaneously. This
is excluded by ∀n ∈ ℕ: 1/n - 1/(n+1) > 0. Note the universal quantifier.
Regards, WM
On 12/3/23 7:44 AM, WM wrote:
Richard Damon schrieb am Sonntag, 3. Dezember 2023 um 13:33:22 UTC+1:
On 12/3/23 6:38 AM, WM wrote:
On 02.12.2023 16:45, Richard Damon wrote:
you still have ℵo
natural numbers you can use,
You cannot use them, because every attempt you make will leave them
unused.
WHy do you say that. What number can't be used? Yes, any number I choose leaves ℵo numbers still above it, but all of them can be used.
No, the laws of mathematics applies to the domain it was derived for.
It
is a fundamental rule that as you expand the set of "Numbers" you are
dealing with, there is a possibility that some of the properties you are
used to fail to hold for those new numbers.
For instance, associativity does NOT hold for infinite sequences, even
those that converge to finite sums, while for finite sequences of terms,
the sum is constant regardless of the order you add the terms.
Because it is mathematical fact. What is at the left-hand side of x is
also at the left-hand side of the interval [x, oo).
To deny this is non-mathematical nonsense.
Yes, and that includes the number x/2 which is also > 0 but in that
interval (0, x), so you don't need to move the numbers to the left of 0.
Your additional unit fractions lie in the interval oflong
(0, x), which has room for an infinite number of unit fractions as
as x > 0.
No.
Then where is x/2?
Your unit fractions are "dense" near 0, which means there is
no "first" unit fraction.
They are not dense enough to avoid the finite intervals between them.
They are.
They are linearly ordered which means that there is a first one after
zero. The only alternative would be many first ones simultaneously.
This is excluded by ∀n ∈ ℕ: 1/n - 1/(n+1) > 0. Note the universal
quantifier.
Nope. try to name it and I can show a number smaller.
They are DENSE about 0, thus, there is no "first"
On 03.12.2023 14:27, Richard Damon wrote:
On 12/3/23 7:44 AM, WM wrote:
Richard Damon schrieb am Sonntag, 3. Dezember 2023 um 13:33:22 UTC+1:
On 12/3/23 6:38 AM, WM wrote:
On 02.12.2023 16:45, Richard Damon wrote:
you still have ℵo
natural numbers you can use,
You cannot use them, because every attempt you make will leave them
unused.
WHy do you say that. What number can't be used? Yes, any number I
choose leaves ℵo numbers still above it, but all of them can be used.
Simply wrong! Finitely many are used. ℵo can never be used: Almost all cannot be used.
No, the laws of mathematics applies to the domain it was derived for.
So it is. All smaller than x is smaller than (x, oo).
∀n ∈ ℕ: 1/n - 1/(n+1) > 0 , Bothe hold foe all numbers.
It is a fundamental rule that as you expand the set of "Numbers" you
are dealing with, there is a possibility that some of the properties
you are used to fail to hold for those new numbers.
I do not consider new numbers but natural numbers which are all finite.
That proves that the new numbers are nonsense.
For instance, associativity does NOT hold for infinite sequences, even
those that converge to finite sums, while for finite sequences of
terms, the sum is constant regardless of the order you add the terms.
All natural numbers are finite and obey the laws of finite numbers.
Never two or more fractions sit at the same point. After "no fraction"
there must be a first fraction and a finite distance to the next
fraction. Simply by the laws of correct mathematics.
Because it is mathematical fact. What is at the left-hand side of x
is also at the left-hand side of the interval [x, oo).
To deny this is non-mathematical nonsense.
Yes, and that includes the number x/2 which is also > 0 but in that
interval (0, x), so you don't need to move the numbers to the left of 0.
Every definable x > 0 is not every x > 0. Every definable x > 0 is
usually named every epsilon > 0. Of course there are ℵo unit fractions smaller than every eps > 0. That proves that you cannot define every x > 0.
Your additional unit fractions lie in the interval oflong
(0, x), which has room for an infinite number of unit fractions as
as x > 0.
No.
Then where is x/2?
"Every x > 0" means all positive points of (0, oo). Here x is not a
certain number, but covers all those points. Therefore also all x/2 are covered.
Your unit fractions are "dense" near 0, which means there is
no "first" unit fraction.
They are not dense enough to avoid the finite intervals between them.
They are.
Either you accept mathematics, in particular ∀n ∈ ℕ: 1/n - 1/(n+1) > 0 or you don't.
They are linearly ordered which means that there is a first one after
zero. The only alternative would be many first ones simultaneously.
This is excluded by ∀n ∈ ℕ: 1/n - 1/(n+1) > 0. Note the universal
quantifier.
Nope. try to name it and I can show a number smaller.
They cannot be named because they are dark.
They are DENSE about 0, thus, there is no "first"
Natural numbers are different, and so are the unit fractions. Dense
would mean that between any two there is another one. That is not true
for unit fractions.
Regards, WM
Richard Damon schrieb am Sonntag, 3. Dezember 2023 um 20:15:21 UTC+1:
On 12/3/23 9:50 AM, WM wrote:
Finitely many are used. ℵo can never be used: Almost allNo, you need to do the operation for ALL natural numbers,
cannot be used.
Yes, they all can be used collectively
|ℕ \ {1, 2, 3, ...}| = 0
But not individually
∀n ∈ ℕ_def: |ℕ \ {1, 2, 3, ..., n}| = ℵo.
That’s why almost all are dark.
Thus you need to use ℵo numbers, and they all
are used.
Yes, but only collectively.
So it is. All smaller than x is smaller than (x, oo). (*)
∀n ∈ ℕ: 1/n - 1/(n+1) > 0. (**)
Both hold for all numbers.
Right, so in the interval (0, x) there are ℵo unit fractions, so no need >> to put them < 0.
All unit fractions which smaller than every x lie at the left-hand side of (x, oo).
This also means that there is no "smallest" unit
fraction to be a "first" point of the interval (0, 1]
Above you seemed to have undersood both (*) and (**)
The function Number of Unit Fractions between 0 and x, NUF(x) is zero at x = 0 and cannot increase to more than 1 without resting at a gap > 0 before the second unit fraction. (**) excludes an increase by more than 1 at any x.
I do not consider new numbers but natural numbers which are all finite.Then why did you comment on trans-finite numbers?
That proves that the new numbers are nonsense.
The following is the reason: Cantor’s bijections are nonsense because of the existzence of dark elements in every infinite set. But we can define | ℕ| or |ℚ| etc.
NOte, once you include "All" Natural Numbers in an operation, you need
to know how to deal with trans-finite operations,
Yes, but not without maintaining logic.|ℚ| > | ℕ|.
By moving the X of the first column in any arbitrary way the matrix
XOOO...
XOOO...
XOOO...
XOOO...
...
Cannot be covered by X.
so even though the
Natural Numbers are all finite, some operations (FOL or HOL) on them
generate infinities.
All natural numbers are finite and obey the laws of finite numbers.Nope, the unit fractions are "dense" about 0,
Never two or more fractions sit at the same point. After "no fraction"
there must be a first fraction and a finite distance to the next
fraction. Simply by the laws of correct mathematics.
No, for all oftem mathematics prescribes
∀n ∈ ℕ: 1/n - 1/(n+1) = 1/(n(n+1)) > 0.
There is nothing “dense”
and thus there is no
"first" fraction, just as there is no "highest" Natural Number.
This shows that there is a last natural number before ω. What else should be there?
Again, while individual Natural Numbers are finite, once you do an
operation that can refer to unbounded sets of them, you need to use
logic that can handle infinites.
Yes, but you seem to forget that. After NUF(x) = 0 for all x =< 0, the increase to NUF(x) = ℵo can only happen in steps of height 1.
not "Definable"? Why are there not ℵo "Definable" Natural Numbers or
Every definable x > 0 is not every x > 0. Every definable x > 0 is
usually named every epsilon > 0. Of course there are ℵo unit fractions >>> smaller than every eps > 0. That proves that you cannot define every x > 0. >> What you mean by "Definable"? What Natural Number (or unit fraction) is
Unit Fractions.
See above. Derfinable as individuals are only finitely many whereas almost all remain dark. But they can be uses collectively:
|ℕ \ {1, 2, 3, ...}| = 0
Try to empty the set by subtracting individually |ℕ \ {1, 2, 3, ..., n}|
Either you accept mathematics, in particular ∀n ∈ ℕ: 1/n - 1/(n+1) > 0How does that say that they are not dense.
or you don't.
Because they all have distances.
Given ANY finite width you
want, you can find an n big enough to put them that close, thus the
field is DENSE at that point.
That is not the meaning of dense. And it further any finite width is large enough to contain almost all unit fractions.
DENSE doesn't mean that you get to non-finite spacing, just that there
doesn't exist a smallest finite spacing, just as there is no largest
finite number.
∀n ∈ ℕ: 1/n - 1/(n+1) = d_n > 0 holds for all distances.
But they can be. Give me a number and I will name the next one.Nope. try to name it and I can show a number smaller.
They cannot be named because they are dark.
Those which can be given are visible. A potentially infinite collection.
Regards, WM
On 12/4/23 4:14 AM, Heinrich wrote:
Richard Damon schrieb am Sonntag, 3. Dezember 2023 um 20:15:21 UTC+1:
On 12/3/23 9:50 AM, WM wrote:
Finitely many are used. ℵo can never be used: Almost allNo, you need to do the operation for ALL natural numbers,
cannot be used.
Yes, they all can be used collectively
|ℕ \ {1, 2, 3, ...}| = 0
But not individually
∀n ∈ ℕ_def: |ℕ \ {1, 2, 3, ..., n}| = ℵo.
That’s why almost all are dark.
But which ones CAN'T be used individually?
Thus you need to use ℵo numbers, and they all
are used.
Yes, but only collectively.
No, they all CAN be used.
All unit fractions which are smaller than every x lie at the left-hand
side of (x, oo).
Yes, and to the right of 0. Remember x > 0, so there is room between it
and 0.
Remember x > 0, so there is room between it
and 0.
Right, but it can be ARBITRARY small. Thus, there is no "first" unit fraction.
Nope. You don't understand the actual meaning of the size of INFINTE
sets, some of the rules of finite sets sizes can't apply,
By moving the X of the first column in any arbitrary way the matrix
XOOO...
XOOO...
XOOO...
XOOO...
...
Cannot be covered by X.
Nope, the unit fractions are "dense" about 0,
No, for all oftem mathematics prescribes
∀n ∈ ℕ: 1/n - 1/(n+1) = 1/(n(n+1)) > 0.
There is nothing “dense”
So, you don't understand what "dense" means.
See above. Derfinable as individuals are only finitely many whereas
almost all remain dark. But they can be uses collectively:
Why only finitely many available? What number is that?
|ℕ \ {1, 2, 3, ...}| = 0
Try to empty the set by subtracting individually |ℕ \ {1, 2, 3, ..., n}|
Just need to do it ℵo times, since I have that many numbers to use.
Either you accept mathematics, in particular ∀n ∈ ℕ: 1/n - 1/(n+1) > 0How does that say that they are not dense.
or you don't.
Because they all have distances.
That get arbitrarily small.
Right, and d_n can get arbitrarily small,
You seem to be confusing HAVE been named, with CAN be named.
Your "dark" seems to actually be a property of the observer, not the
numbers.
On 05.12.2023 01:09, Richard Damon wrote:
On 12/4/23 4:14 AM, Heinrich wrote:
Richard Damon schrieb am Sonntag, 3. Dezember 2023 um 20:15:21 UTC+1:
On 12/3/23 9:50 AM, WM wrote:
Finitely many are used. ℵo can never be used: Almost allNo, you need to do the operation for ALL natural numbers,
cannot be used.
Yes, they all can be used collectively
|ℕ \ {1, 2, 3, ...}| = 0
But not individually
∀n ∈ ℕ_def: |ℕ \ {1, 2, 3, ..., n}| = ℵo.
That’s why almost all are dark.
But which ones CAN'T be used individually?
Try it. Take the largest n and remove {1, 2, 3, ..., n} from ℕ. Look whether something remains.
Thus you need to use ℵo numbers, and they all
are used.
Yes, but only collectively.
No, they all CAN be used.
Try it. Take the largest n and remove {1, 2, 3, ..., n} from ℕ. Look whether something remains.
All unit fractions which are smaller than every x lie at the left-hand
side of (x, oo).
Yes, and to the right of 0. Remember x > 0, so there is room between it and 0.
There is no room between 0 and (0, oo).
Remember x > 0, so there is room between it and 0.
Remember: All x > 0 with no exception.
Right, but it can be ARBITRARY small. Thus, there is no "first" unit
fraction.
Arbitrary small is finite. All unit fractions are different and sit at different places. So many first ones together is impossible, hence there
is a single first one.
Nope. You don't understand the actual meaning of the size of INFINTE
sets, some of the rules of finite sets sizes can't apply,
I understand that many people are too dense to understand that they have
been fooled.
By moving the X of the first column in any arbitrary way the matrix
XOOO...
XOOO...
XOOO...
XOOO...
...
Cannot be covered by X.
Nope, the unit fractions are "dense" about 0,
No, for all oftem mathematics prescribes
∀n ∈ ℕ: 1/n - 1/(n+1) = 1/(n(n+1)) > 0.
There is nothing “dense”
So, you don't understand what "dense" means.
I have learnt and I teach that it means: Between any two pints there is another point. That is not true for unit fractions because between 1/n
and 1/(n+1) there is no further unit fraction. Perhaps your mistake is
based on your being wrong in this respect.
See above. Derfinable as individuals are only finitely many whereas
almost all remain dark. But they can be uses collectively:
Why only finitely many available? What number is that?
The set is potentially infinite but never as large as the set of dark numbers.
|ℕ \ {1, 2, 3, ...}| = 0Just need to do it ℵo times, since I have that many numbers to use.
Try to empty the set by subtracting individually |ℕ \ {1, 2, 3, ..., n}| >>
No that is naive and wrong. You cannot exhaust the set ℕ. The not used numbers will always remain |ℕ \ ℕ_vis| = ℵo. No chance for you.
Either you accept mathematics, in particular ∀n ∈ ℕ: 1/n - 1/(n+1) > 0How does that say that they are not dense.
or you don't.
Because they all have distances.
That get arbitrarily small.
Nevertheless they remain distances.
Right, and d_n can get arbitrarily small,
Nevertheless it remains a distance.
You seem to be confusing HAVE been named, with CAN be named.
|ℕ \ ℕ_vis| = ℵo. Therefore most numbers will never be named and cannot be named.
Your "dark" seems to actually be a property of the observer, not the
numbers.
Yes, that is true. But |ℕ \ ℕ_vis| = ℵo cannot be undercut.
Regards, WM
On 12/5/23 5:49 AM, WM wrote:
On 05.12.2023 01:09, Richard Damon wrote:
On 12/4/23 4:14 AM, Heinrich wrote:
Richard Damon schrieb am Sonntag, 3. Dezember 2023 um 20:15:21 UTC+1: >> >>> On 12/3/23 9:50 AM, WM wrote:;
;Finitely many are used. ℵo can never be used: Almost allNo, you need to do the operation for ALL natural numbers,
cannot be used.
Yes, they all can be used collectively
|ℕ \ {1, 2, 3, ...}| = 0
But not individually
∀n ∈ ℕ_def: |ℕ \ {1, 2, 3, ..., n}| = ℵo.
That’s why almost all are dark.
But which ones CAN'T be used individually?
Try it. Take the largest n and remove {1, 2, 3, ..., n} from ℕ. Look
whether something remains.
WHAT "Largest n". There is no such thing.
There is no room between 0 and (0, oo).
But it isn't between o and (0, oo), but between 0 and (x, 00)
Remember: All x > 0 with no exception.
Right, and for every x, there is a smaller y available.
So, you don't understand what "dense" means.
I have learnt and I teach that it means: Between any two points there
is another point. That is not true for unit fractions because between
1/n and 1/(n+1) there is no further unit fraction. Perhaps your
mistake is based on your being wrong in this respect.
See above. Derfinable as individuals are only finitely many whereas
almost all remain dark. But they can be uses collectively:
Why only finitely many available? What number is that?
The set is potentially infinite but never as large as the set of dark
numbers.
But it is, you just don't understand how it gets there.
Yes, you CAN exhaust the Natural Numbers if you do something ℵo times.
If you can't do something ℵo times, you can't have the Natural Numbers
in the first place.
Right, and d_n can get arbitrarily small,
Nevertheless it remains a distance.
But as small as we want.
You seem to be confusing HAVE been named, with CAN be named.
|ℕ \ ℕ_vis| = ℵo. Therefore most numbers will never be named and
cannot be named.
Only if N_vis is restricted to be bounded.
Your "dark" seems to actually be a property of the observer, not the
numbers.
Yes, that is true. But |ℕ \ ℕ_vis| = ℵo cannot be undercut.
Nope, Since ℕ_vis is actually ℕ, since all Natural Numbers are visible
Your ultimate problem is that you are trying to talk about an infinite
set with operations that can only apply to finite sets. Your logic
system is just incabable of handling infinities, because it is just do
small.
Your "dark numbers" are just the Natural Numbers that you logic system, because it is too weak to actually have the infinite number of Natural Numbers, doesn't support. They are the gap between the bounded set your
logic can handle, and the unbounded set of the Natural Numbers.
On 05.12.2023 13:23, Richard Damon wrote:
WHAT "Largest n". There is no such thing.
Take the largest you can take as an individual.
If it turns out not be the largest but
is surpassed by another one,
take the larger one.
Continue as long as it satisfies you.
Never you will have taken all.
Remember: All x > 0 with no exception.
Right,
and for every x,
there is a smaller y available.
If you remove all x > 0 from the real axis
then nevertheless positive points remain?
On 05.12.2023 13:23, Richard Damon wrote:
On 12/5/23 5:49 AM, WM wrote:
On 05.12.2023 01:09, Richard Damon wrote:
On 12/4/23 4:14 AM, Heinrich wrote:UTC+1:
Richard Damon schrieb am Sonntag, 3. Dezember 2023 um 20:15:21
;On 12/3/23 9:50 AM, WM wrote:;
Finitely many are used. ℵo can never be used: Almost allNo, you need to do the operation for ALL natural numbers,
cannot be used.
Yes, they all can be used collectively
|ℕ \ {1, 2, 3, ...}| = 0
But not individually
∀n ∈ ℕ_def: |ℕ \ {1, 2, 3, ..., n}| = ℵo.
That’s why almost all are dark.
But which ones CAN'T be used individually?
Try it. Take the largest n and remove {1, 2, 3, ..., n} from ℕ. Look
whether something remains.
WHAT "Largest n". There is no such thing.
Take the largest you can take as an individual. If it turns out not be
the largest but is surpassed by another one, take the larger one.
Continue as long as it satisfies you. Never you will have taken all.
There is no room between 0 and (0, oo).
But it isn't between o and (0, oo), but between 0 and (x, 00)
(0, oo) contains all x such there no positive point is left. There is no
room in between.
Remember: All x > 0 with no exception.
Right, and for every x, there is a smaller y available.
If you remove all x > 0 from the real axis then nevertheless positive
points remain?
Dream on.
So, you don't understand what "dense" means.
I have learnt and I teach that it means: Between any two points
there is another point. That is not true for unit fractions because
between 1/n and 1/(n+1) there is no further unit fraction. Perhaps
your mistake is based on your being wrong in this respect.
No answer?
See above. Derfinable as individuals are only finitely many whereas
almost all remain dark. But they can be uses collectively:
Why only finitely many available? What number is that?
The set is potentially infinite but never as large as the set of dark
numbers.
But it is, you just don't understand how it gets there.
It does not get there. You cannot exhaust the set ℕ. The not used
numbers will always remain |ℕ \ ℕ_vis| = ℵo. No chance for you.
Yes, you CAN exhaust the Natural Numbers if you do something ℵo times.
No, that is nonsense. What will be the last number?
If you can't do something ℵo times, you can't have the Natural Numbers
in the first place.
You can treat them all collectively but not individually.
Right, and d_n can get arbitrarily small,
Nevertheless it remains a distance.
But as small as we want.
No, you want zero. That is impossible, Therefore there is a first step
of NUF(x) with a tiny stop before the next one.
You seem to be confusing HAVE been named, with CAN be named.
|ℕ \ ℕ_vis| = ℵo. Therefore most numbers will never be named and
cannot be named.
Only if N_vis is restricted to be bounded.
Not bounded by any finite number. But bounded by the condition that ℵo numbers will remain.
Your "dark" seems to actually be a property of the observer, not the
numbers.
Yes, that is true. But |ℕ \ ℕ_vis| = ℵo cannot be undercut.
Nope, Since ℕ_vis is actually ℕ, since all Natural Numbers are visible
Then show the last one.
Your ultimate problem is that you are trying to talk about an infinite
set with operations that can only apply to finite sets. Your logic
system is just incabable of handling infinities, because it is just do
small.
It is correct, contrary to yours.
Your "dark numbers" are just the Natural Numbers that you logic
system, because it is too weak to actually have the infinite number of
Natural Numbers, doesn't support. They are the gap between the bounded
set your logic can handle, and the unbounded set of the Natural Numbers.
They are the gap. You cannot bridge it because every number visible to
you has ℵo successors which are numbers too but cannot be removed individually.
Regards, WM
On 12/6/2023 12:24 PM, WM wrote:
On 05.12.2023 13:23, Richard Damon wrote:
WHAT "Largest n". There is no such thing.
Take the largest you can take as an individual.
ℒ is the largest natural number which
I can take as an individual.
I know that
ℒ is not the largest natural number.
ℒ is surpassed by ℒ+1
If you remove all x > 0 from the real axis
then nevertheless positive points remain?
You (WM) are using a quantifier shift, and
"forgetting" to mention that you are.
Mention or not-mention.
Quantifier shift is not reliably not-first-false.
Not-mentioning won't make you correct.
It will only make it harder to see
why you're incorrect.
Making it harder to see why you're incorrect
seems to be what your rhetoric is designed to do.
Consider being correct, instead.
On 12/6/23 12:24 PM, WM wrote:
Take the largest you can take as an individual. If it turns out not be
the largest but is surpassed by another one, take the larger one.
Continue as long as it satisfies you. Never you will have taken all.
Which is bounded logic that can't create the Natural Numbers,
(0, oo) contains all x such there no positive point is left. There is
no room in between.
Yes, but there is not first point in (0, oo) so that doesn't matter.
Remember: All x > 0 with no exception.
Right, and for every x, there is a smaller y available.
If you remove all x > 0 from the real axis then nevertheless positive
points remain?
Dream on.
I didn't say that, you did.
So, you don't understand what "dense" means.
I have learnt and I teach that it means: Between any two points
there is another point. That is not true for unit fractions because
between 1/n and 1/(n+1) there is no further unit fraction. Perhaps
your mistake is based on your being wrong in this respect.
No answer?
What need is there to be a fraction between 1/n and 1/(n+1).
The claim is that there is always a faction between 1/n and 0, since no 1/(n+1) is == 0, those are diffferent statements.
The "infinite" set of the "describable" Natural Numbers is EXACTLY the
same set as the Natural Numbers, so when you remove them, you get no
number left to be "dark".
Nope, the unbounded n bridges the gap. Yes, it takes infinite "steps" to
get there,
On 06.12.2023 19:44, Jim Burns wrote:
[...]
That makes potentiel infinity
a difficult concept.
But it can be handled by your knowledge that
for every number that can
be taken there are ℵo greater numbers,
ℵo of which can never be taken.
On 06.12.2023 19:44, Jim Burns wrote:
On 12/6/2023 12:24 PM, WM wrote:
On 05.12.2023 13:23, Richard Damon wrote:
WHAT "Largest n". There is no such thing.
Take the largest you can take as an individual.
ℒ is the largest natural number which
I can take as an individual.
I know that
ℒ is not the largest natural number.
ℒ is surpassed by ℒ+1
That makes potentiel infinity
a difficult concept.
On 12/7/2023 10:47 AM, WM wrote:
That makes potentiel infinity
a difficult concept.
Your potential.infinityᵂᴹ wanders
between being about numbers I can take
and about numbers which exist.
On 07.12.2023 02:07, Richard Damon wrote:
On 12/6/23 12:24 PM, WM wrote:
Take the largest you can take as an individual. If it turns out not be
the largest but is surpassed by another one, take the larger one.
Continue as long as it satisfies you. Never you will have taken all.
Which is bounded logic that can't create the Natural Numbers,
It is all that you can do.
(0, oo) contains all x such there no positive point is left. There is
no room in between.
Yes, but there is not first point in (0, oo) so that doesn't matter.
There is a first point. But that is not esil proven. However there is a
first unit fraction, easily proven by ∀n ∈ ℕ: 1/n - 1/(n+1) = d_n > 0.
Remember: All x > 0 with no exception.
Right, and for every x, there is a smaller y available.
If you remove all x > 0 from the real axis then nevertheless positive
points remain?
Dream on.
I didn't say that, you did.
RD: Right, and for every x, there is a smaller y available.
So, you don't understand what "dense" means.
I have learnt and I teach that it means: Between any two points
there is another point. That is not true for unit fractions because
between 1/n and 1/(n+1) there is no further unit fraction. Perhaps
your mistake is based on your being wrong in this respect.
No answer?
What need is there to be a fraction between 1/n and 1/(n+1).
The claim is that there is always a faction between 1/n and 0, since no 1/(n+1) is == 0, those are diffferent statements.
That is not the meaning of dense.
The "infinite" set of the "describable" Natural Numbers is EXACTLY the same set as the Natural Numbers, so when you remove them, you get no number left to be "dark".
That is contradicted by the fact that describable means possible to be removed as an individual, but then ∀n ∈ ℕ_def: |ℕ \ {1, 2, 3, ..., n}| =
ℵo.
Nope, the unbounded n bridges the gap. Yes, it takes infinite "steps" to get there,
No. For every step of all infinitely many steps |ℕ \ {1, 2, 3, ..., n}|
= ℵo remains true.
If you can't understandd that, then EOD.
Regards, WM
On 08.12.2023 05:09, Jim Burns wrote:
On 12/7/2023 10:47 AM, WM wrote:
That makes potentiel infinity
a difficult concept.
Your potential.infinityᵂᴹ wanders
between being about numbers I can take
and about numbers which exist.
With n also n+1 is visible.
But always almost all numbers are dark.
I can't change that.
I can only hint to the unit fractions.
There are none smaller than 0.
So all are restricted to the positive axis.
But the first ℵo unit fractions cannot be seen.
Although you can see
for every unit fraction a smaller one,
the visible unit fractions are always
a finite set.
On 12/8/2023 10:21 AM, WM wrote:
On 08.12.2023 05:09, Jim Burns wrote:
On 12/7/2023 10:47 AM, WM wrote:
That makes potentiel infinity
a difficult concept.
Your potential.infinityᵂᴹ wanders
between being about numbers I can take
and about numbers which exist.
With n also n+1 is visible.
That matters. Quite a lot.
We can say _once_ and
know _infinitely.many ways_ that
n+1 in ⟨0,…,n,n+1⟩ is not.first.darkᵂᴹ
We can follow that by saying once and
knowing infinitely.many ways that
n+1 in ⟨0,…,n,n+1⟩ is visibleᵂᴹ
We can say once and
know infinitely.many ways that
unit fraction ⅟n is not-in the first ℵ₀
The first ℵ₀ is empty.
And empty is not ℵ₀-many.
Contradiction.
Visibleᵂᴹ unit fractions such that
for every unit fraction there's a smaller one,
are not 1.by.1;2.end.able.
Thus, they are not a finite set.
On 12/7/23 11:10 AM, WM wrote:
There is a first point. But that is not easily proven. However there is
a first unit fraction, easily proven by ∀n ∈ ℕ: 1/n - 1/(n+1) = d_n > 0.
Nope, there is no first point, because that number system is dense
there.
Between 0 and ANY point, even your "claimed first" point, there
are more points, therefore, there can't be a "first point"
On 08.12.2023 20:46, Richard Damon wrote:
On 12/7/23 11:10 AM, WM wrote:
There is a first point. But that is not easily proven. However there
is a first unit fraction, easily proven by ∀n ∈ ℕ: 1/n - 1/(n+1) = >>> d_n > 0.
Nope, there is no first point, because that number system is dense there.
I told you already that the unit fractions, contrary to the fractions,
have nothing to do with dense.
Between 0 and ANY point, even your "claimed first" point, there are
more points, therefore, there can't be a "first point"
If you deny mathematics, then you can claim whatever you like. Nobody
can contradict your belief. In mathematics we have
∀n ∈ ℕ: 1/n - 1/(n+1) = d_n > 0
with a universal quantifier. No unit fraction can violate this statement.
Regards, WM
On 12/9/23 6:17 AM, WM wrote:
On 08.12.2023 20:46, Richard Damon wrote:
On 12/7/23 11:10 AM, WM wrote:
There is a first point. But that is not easily proven. However there
is a first unit fraction, easily proven by ∀n ∈ ℕ: 1/n - 1/(n+1) = >>>> d_n > 0.
Nope, there is no first point, because that number system is dense
there.
I told you already that the unit fractions, contrary to the fractions,
have nothing to do with dense.
Which just shows you don't know what it means, or are a liar.
∀n ∈ ℕ: 1/n - 1/(n+1) = d_n > 0
with a universal quantifier. No unit fraction can violate this statement.
So?, that isn't saying that there isn't a first point.
There is no finite distance x
On 09.12.2023 13:27, Richard Damon wrote:
On 12/9/23 6:17 AM, WM wrote:
On 08.12.2023 20:46, Richard Damon wrote:
On 12/7/23 11:10 AM, WM wrote:
There is a first point. But that is not easily proven. However
there is a first unit fraction, easily proven by ∀n ∈ ℕ: 1/n - >>>>> 1/(n+1) = d_n > 0.
Nope, there is no first point, because that number system is dense
there.
I told you already that the unit fractions, contrary to the
fractions, have nothing to do with dense.
Which just shows you don't know what it means, or are a liar.
From Wikipedia, the free encyclopedia
"In mathematics, a partial order or total order < on a set X is said to
be dense if, for all x and y in X for which x < y, there is a z in X
such that x < z < y." Note for all. You could have looked up that yourself.
∀n ∈ ℕ: 1/n - 1/(n+1) = d_n > 0
with a universal quantifier. No unit fraction can violate this
statement.
So?, that isn't saying that there isn't a first point.
No. There is a first point.
There is no finite distance x
There are finite distances between all unit fractions. Of course we
cannot get them because they are dark.
Regards, WM
On 08.12.2023 21:37, Jim Burns wrote:
On 12/8/2023 10:21 AM, WM wrote:
On 08.12.2023 05:09, Jim Burns wrote:
On 12/7/2023 10:47 AM, WM wrote:
That makes potentiel infinity
a difficult concept.
Your potential.infinityᵂᴹ wanders
between being about numbers I can take
and about numbers which exist.
With n also n+1 is visible.
That matters. Quite a lot.
We can say _once_ and
know _infinitely.many ways_ that
n+1 in ⟨0,…,n,n+1⟩ is not.first.darkᵂᴹ
We can follow that by saying once and
knowing infinitely.many ways that
n+1 in ⟨0,…,n,n+1⟩ is visibleᵂᴹ
Yes.
Nevertheless
∀n ∈ ℕ_vis:
|ℕ \ {1, 2, 3, ..., n}| = ℵo .
https://en.wikipedia.org/wiki/Programme_for_International_Student_AssessmentWe can say once and
know infinitely.many ways that
unit fraction ⅟n is not-in the first ℵ₀
The first ℵ₀ is empty.
And empty is not ℵ₀-many.
Contradiction.
If your logic says that
there is no smallest unit fraction,
then it is in contradiction with
mathematics which even a Pisa-pupil
should master:
∀n ∈ ℕ: 1/n - 1/(n+1) = d_n > 0 .
Note the universal quantifier which
does not admit ***any*** exception.
On 12/9/23 11:11 AM, WM wrote:
From Wikipedia, the free encyclopedia
"In mathematics, a partial order or total order < on a set X is said
to be dense if, for all x and y in X for which x < y, there is a z in
X such that x < z < y." Note for all. You could have looked up that
yourself.
Yes, and the extension of that definition, for Dense at a point,
restricts that x or y is that point. In this case, x, which is 0.
∀n ∈ ℕ: 1/n - 1/(n+1) = d_n > 0
with a universal quantifier. No unit fraction can violate this
statement.
So?, that isn't saying that there isn't a first point.
No. There is a first point.
Then what is it?
The distance is finite, but unboundedly small, thus no first element,
which would become a bound.
On 12/9/2023 6:11 AM, WM wrote:
On 08.12.2023 21:37, Jim Burns wrote:
We can follow that by saying once and
knowing infinitely.many ways that
n+1 in ⟨0,…,n,n+1⟩ is visibleᵂᴹ
Yes.
Nevertheless
∀n ∈ ℕ_vis:
|ℕ \ {1, 2, 3, ..., n}| = ℵo .
Yes.
|ℕ\{1,2,3,...,n}| = ℵ₀
means
|ℕ| = ℵ₀ = |ℤ| though ℕ ≠⊂ ℤ
ℕ is a same "sized" proper subset of ℤ
Same "sized" proper subsets
If your logic says thathttps://en.wikipedia.org/wiki/Programme_for_International_Student_Assessment
there is no smallest unit fraction,
then it is in contradiction with
mathematics which even a Pisa-pupil
should master:
∀n ∈ ℕ: 1/n - 1/(n+1) = d_n > 0 .
We can say once
1/n - 1/(n+1) = d_n > 0
and know it is true in infinitely.many ways
for each number in ℕ
⅟n - ⅟n⁺¹ > 0
⅟n⁺¹ < ⅟n
⅟n is not the first unit fraction
We can say once
⅟n is not the first unit fraction
and know it is true in infinitely.many ways
for each unit fraction in ⅟ℕ
Note the universal quantifier which
does not admit ***any*** exception.
We can say once
⅟n is not an exception to being not.first in ⅟ℕ
and know it is true in infinitely-many ways
for each unit fraction in ⅟ℕ
On 09.12.2023 17:32, Richard Damon wrote:
On 12/9/23 11:11 AM, WM wrote:
From Wikipedia, the free encyclopedia
"In mathematics, a partial order or total order < on a set X is said
to be dense if, for all x and y in X for which x < y, there is a z in
X such that x < z < y." Note for all. You could have looked up that
yourself.
Yes, and the extension of that definition, for Dense at a point,
restricts that x or y is that point. In this case, x, which is 0.
That is not an extension but a wrong application.
∀n ∈ ℕ: 1/n - 1/(n+1) = d_n > 0
with a universal quantifier. No unit fraction can violate this
statement.
So?, that isn't saying that there isn't a first point.
You are right
No. There is a first point.
Then what is it?
It is dark.
The distance is finite, but unboundedly small, thus no first element,
which would become a bound.
That doesn't make it disappear. Mathematics needs precision, not the
belief that unboundedly small is same as vanishing.
Regards, WM
On 09.12.2023 20:56, Jim Burns wrote:
On 12/9/2023 6:11 AM, WM wrote:
On 08.12.2023 21:37, Jim Burns wrote:
We can follow that by saying once and
knowing infinitely.many ways that
n+1 in ⟨0,…,n,n+1⟩ is visibleᵂᴹ
Yes.
Nevertheless
∀n ∈ ℕ_vis:
|ℕ \ {1, 2, 3, ..., n}| = ℵo .
Yes.
|ℕ\{1,2,3,...,n}| = ℵ₀
means
that almost all natnumbers are dark.
If your logic says thathttps://en.wikipedia.org/wiki/Programme_for_International_Student_Assessment >>> should master:
there is no smallest unit fraction,
then it is in contradiction with
mathematics which even a Pisa-pupil
∀n ∈ ℕ: 1/n - 1/(n+1) = d_n > 0 .
On 12/10/23 3:53 AM, WM wrote:
On 09.12.2023 17:32, Richard Damon wrote:
On 12/9/23 11:11 AM, WM wrote:
From Wikipedia, the free encyclopedia
"In mathematics, a partial order or total order < on a set X is said
to be dense if, for all x and y in X for which x < y, there is a z
in X such that x < z < y." Note for all. You could have looked up
that yourself.
Yes, and the extension of that definition, for Dense at a point,
restricts that x or y is that point. In this case, x, which is 0.
That is not an extension but a wrong application.
So, you don't understand how logic works.
∀n ∈ ℕ: 1/n - 1/(n+1) = d_n > 0
with a universal quantifier. No unit fraction can violate this
statement.
So?, that isn't saying that there isn't a first point.
You are right
No. There is a first point.
Then what is it?
It is dark.
Nope, you just agreed that there wasn't one, so it can't be dark.
The distance is finite, but unboundedly small, thus no first element,
which would become a bound.
That doesn't make it disappear. Mathematics needs precision, not the
belief that unboundedly small is same as vanishing.
Why? What is the difference between the terms?
On 12/10/2023 5:13 AM, WM wrote:
|ℕ\{1,2,3,...,n}| = ℵ₀
means
that almost all natnumbers are dark.
That's not what ℵ₀ means.
ℵ₀.many are 1.by.1;1.end.able.
However,
what you (WM) mean is
∀n ∈ ℕ_vis:
almost all in ℕᵂᴹ\{1,2,3,...,n} are darkᵂᴹ
Elsewhere, you:
| With n also n+1 is visible.
|
If your logic says thathttps://en.wikipedia.org/wiki/Programme_for_International_Student_Assessment
there is no smallest unit fraction,
then it is in contradiction with
mathematics which even a Pisa-pupil
should master:
∀n ∈ ℕ: 1/n - 1/(n+1) = d_n > 0 .
Imagine how shocked I am to learn that
it is _you_ Wolfgang Mückenheim who is contradicting
the mathematics a PISA student should master.
On 10.12.2023 14:07, Richard Damon wrote:
On 12/10/23 3:53 AM, WM wrote:
On 09.12.2023 17:32, Richard Damon wrote:
On 12/9/23 11:11 AM, WM wrote:
From Wikipedia, the free encyclopedia
"In mathematics, a partial order or total order < on a set X is
said to be dense if, for all x and y in X for which x < y, there is
a z in X such that x < z < y." Note for all. You could have looked
up that yourself.
Yes, and the extension of that definition, for Dense at a point,
restricts that x or y is that point. In this case, x, which is 0.
That is not an extension but a wrong application.
So, you don't understand how logic works.
Try to study the meaning of mathematical definitions. Also others here
have pointed out that you are wrong.
∀n ∈ ℕ: 1/n - 1/(n+1) = d_n > 0
with a universal quantifier. No unit fraction can violate this
statement.
So?, that isn't saying that there isn't a first point.
You are right
No. There is a first point.
Then what is it?
It is dark.
Nope, you just agreed that there wasn't one, so it can't be dark.
You said: "that isn't saying that there isn't a first point". That means
that there can be a first point. Of course I agreed.
The distance is finite, but unboundedly small, thus no first
element, which would become a bound.
That doesn't make it disappear. Mathematics needs precision, not the
belief that unboundedly small is same as vanishing.
Why? What is the difference between the terms?
The difference is that between finite and not finite. All natural
numbers, with no exception, are finite.
Regards, WM
On 11.12.2023 05:50, Jim Burns wrote:
[...]
No quantifier magic will help you in this case:
If ℵ₀ points are left-hand side of
every positive point,
then ℵ₀ points are left-hand side of
all positive points.
This proves that ℵ₀ points
are left-hand side of the interval (0, 1].
On 12/11/2023 5:58 AM, WM wrote:
On 11.12.2023 05:50, Jim Burns wrote:
[...]
No quantifier magic will help you in this case:
If ℵ₀ points are left-hand side of
every positive point,
then ℵ₀ points are left-hand side of
all positive points.
... "quantifier magic" in arithmetic which
PISA students must master.
Consider a use of "quantifier magic" for finites.
Four friends are going to have dinner together,
Nora, Sarah, Earl, and William.
I'm setting the table.
I take four glasses, set them at Nora's chair,
and go back into the kitchen to check the roast.
N:1234 S E W
While checking the roast, I tell you
| For each glass 1234
| there is a chair NSEW
| such that
| the chair faces the glass
A bit later,
I go back and distribute the glasses.
N:1 S:2 E:3 W:4
I tell you a second time
| For each glass 1234
| there is a chair NSEW
| such that
| the chair faces the glass
The "quantifier magic" for this table
is to not.conclude that,
| There _is_ a chair NSEW
| such that
| for each glass 1234
| the chair faces the glass
Where there are infinitely.many,
you also must not.conclude.
We know that
| For each unit.fraction ⅟n
| there is an ℵ₀.many.subset S ⁱⁿᶠ⊆ ⅟ℕ
| such that
| S is bounded above by ⅟n: S ᣔ≤ ⅟n
From only that, you must not-conclude that
| There is an ℵ₀.many.subset S ⁱⁿᶠ⊆ ⅟ℕ
| such that,
| for each unit.fraction ⅟n
| S is bounded above by ⅟n: S ᣔ≤ ⅟n
However,
you can conclude the negation of that
| There _isn't_ an ℵ₀.many.subset S ⁱⁿᶠ⊆ ⅟ℕ
| such that,
| for each unit.fraction ⅟n
| S is bounded above by ⅟n: S ᣔ≤ ⅟n
| Assume otherwise.
| Assume
| there _is_ an ℵ₀.many.subset S ⁱⁿᶠ⊆ ⅟ℕ
| such that,
| for each unit.fraction ⅟n
| S is bounded above by ⅟n: S ᣔ≤ ⅟n
|
| However,
| S is not bounded by
| the _second.largest_ unit fraction ⅟mₛ⁺¹ in S
| S is not bounded by each unit.fraction
| Contradiction.
Therefore, there isn't.
This proves that ℵ₀ points are
left-hand side of the interval (0, 1].
Your claim here is equivalent to,
in the glasses-and-chairs example,
thinking of only N:1234 S E W
and
forgetting about N:1 S:2 E:3 W:4
On 12/11/23 5:48 AM, WM wrote:
Try to study the meaning of mathematical definitions. Also others hereAnd which definitions are you trying to use?
have pointed out that you are wrong.
The is nothing in the statement ∀n ∈ ℕ: 1/n - 1/(n+1) = d_n > 0 that even implies that there is or has to be a "first point",
The difference is that between finite and not finite. All naturalYes, but unbounded, which is the bridge to infinite.
numbers, with no exception, are finite.
Every Natural Number is finite, but as a set, they are infinite.
On 12/11/2023 5:58 AM, WM wrote:
This proves that ℵ₀ points are left-hand side of the interval (0, 1].
Your claim here is equivalent to,
in the glasses-and-chairs example,
thinking of only N:1234 S E W
and
forgetting about N:1 S:2 E:3 W:4
Richard Damon schrieb am Montag, 11. Dezember 2023 um 13:19:35 UTC+1:
On 12/11/23 5:48 AM, WM wrote:
Try to study the meaning of mathematical definitions. Also othershere > have pointed out that you are wrong.
And which definitions are you trying to use?
Those of text books like W. Mückenheim: "Mathematik für die ersten Semester", 4th ed., De Gruyter, Berlin (2015) or W. Mückenheim: "Die Mathematik des Unendlichen", Shaker-Verlag, Aachen 2006.
The is nothing in the statement ∀n ∈ ℕ: 1/n - 1/(n+1) = d_n > 0 that >> even implies that there is or has to be a "first point",
It is. You are only too blind to see it. Between all points there is are gaps. And there is no point on the negative axis.
The difference is that between finite and not finite. All natural >numbers, with no exception, are finite.
Yes, but unbounded, which is the bridge to infinite.
Bridge maybe, but no natnumber enters this bridge.
Every Natural Number is finite, but as a set, they are infinite.
We talk about unit fractions, not about the set.
Regards, WM
On 12/12/23 6:57 AM, WM wrote:
There is not least Unit Fraction, because that would imply a maximum
Natural Number, which there isn't one.
Thus, for Unit Fractions, there is no least element, as the set is
Unbounded, just as there is not last Natural Number as that set is
Unbounded in the same way.
On 12.12.2023 13:21, Richard Damon wrote:
On 12/12/23 6:57 AM, WM wrote:
There is not least Unit Fraction, because that would imply a maximum
Natural Number, which there isn't one.
There is NUF(0) = 0 and NUF(x>0) > 0. Never two unit fractions sit at
the same point. This proves a first one.
For natnumbers there is a corresponding last one. They are dark like the smallest unit fractions.
Thus, for Unit Fractions, there is no least element, as the set is
Unbounded, just as there is not last Natural Number as that set is
Unbounded in the same way.
When you step down from omega, you get to a definable natural number n. Between n and omega however, there are ℵo natnumbers. Why can't you step down to any of them? Because they are dark.
Regards, WM
On 12/12/23 5:12 PM, WM wrote:
On 12.12.2023 13:21, Richard Damon wrote:
On 12/12/23 6:57 AM, WM wrote:
There is not least Unit Fraction, because that would imply a maximum
Natural Number, which there isn't one.
There is NUF(0) = 0 and NUF(x>0) > 0. Never two unit fractions sit at
the same point. This proves a first one.
Which "NUF(x)" are you trying to use?
NUF(x) being the number of unit fractions below x is 0 for x <=0 and
infinite for x > 0, so has a major discontinuity at 0
NUF(x) being the number of unit fractions above x
For natnumbers there is a corresponding last one. They are dark like
the smallest unit fractions.
Which is the "Last" Natural Number?
Thus, for Unit Fractions, there is no least element, as the set is
Unbounded, just as there is not last Natural Number as that set is
Unbounded in the same way.
When you step down from omega, you get to a definable natural number
n. Between n and omega however, there are ℵo natnumbers. Why can't you
step down to any of them? Because they are dark.
You CAN'T "Step Down" from omega.
On 13.12.2023 00:46, Richard Damon wrote:
On 12/12/23 5:12 PM, WM wrote:
On 12.12.2023 13:21, Richard Damon wrote:
On 12/12/23 6:57 AM, WM wrote:
There is not least Unit Fraction, because that would imply a maximum
Natural Number, which there isn't one.
There is NUF(0) = 0 and NUF(x>0) > 0. Never two unit fractions sit at
the same point. This proves a first one.
Which "NUF(x)" are you trying to use?
NUF(x) being the number of unit fractions below x is 0 for x <=0 and
infinite for x > 0, so has a major discontinuity at 0
Impossible. Contradicting mathematics: Never two or more unit fractions
sit at the same place.
NUF(x) being the number of unit fractions above x
No, below x. NUF(x) is the number between 0 and x.
For natnumbers there is a corresponding last one. They are dark like
the smallest unit fractions.
Which is the "Last" Natural Number?
Dark. Impossiblr to discern.
Thus, for Unit Fractions, there is no least element, as the set is
Unbounded, just as there is not last Natural Number as that set is
Unbounded in the same way.
When you step down from omega, you get to a definable natural number
n. Between n and omega however, there are ℵo natnumbers. Why can't
you step down to any of them? Because they are dark.
You CAN'T "Step Down" from omega.
It can be done. It is a general rule, forced by the well-foundedness
(see 2.12.8) of the sequence of ordinal numbers, that every strictly decreasing sequence of ordinal numbers reaches its smallest element
after a finite number of steps. Since limit ordinals have no direct predecessors, we have to jump down from them to some predecessor.
Ask for help elsewhere if you don't trust me.
Regards, WM
On 12/13/23 7:01 AM, WM wrote:
On 13.12.2023 00:46, Richard Damon wrote:
NUF(x) being the number of unit fractions below x is 0 for x <=0 and
infinite for x > 0, so has a major discontinuity at 0
Impossible. Contradicting mathematics: Never two or more unit
fractions sit at the same place.
Right, no two at the same space, but there "density" becomes infinite,
and thus we get the discontinuity,
You just don't understand what you need to do to work with infinite sets.
NUF(x) being the number of unit fractions above x
No, below x. NUF(x) is the number between 0 and x.
So, it is INFINITE for all x > 0
So, what is the last discernable number?
You CAN'T "Step Down" from omega.
It can be done. It is a general rule, forced by the well-foundedness
(see 2.12.8) of the sequence of ordinal numbers, that every strictly
decreasing sequence of ordinal numbers reaches its smallest element
after a finite number of steps. Since limit ordinals have no direct
predecessors, we have to jump down from them to some predecessor.
Ask for help elsewhere if you don't trust me.
What is 2.12.8?
Your arguemnt is assuming its conclusion. Omega isn't a Natural Number,
so you can't use arguements about Natural Numbers to talk about it.
On 13.12.2023 13:36, Richard Damon wrote:
On 12/13/23 7:01 AM, WM wrote:
On 13.12.2023 00:46, Richard Damon wrote:
NUF(x) being the number of unit fractions below x is 0 for x <=0 and
infinite for x > 0, so has a major discontinuity at 0
Impossible. Contradicting mathematics: Never two or more unit
fractions sit at the same place.
Right, no two at the same space, but there "density" becomes infinite,
and thus we get the discontinuity,
No. All distances remain finite.
You just don't understand what you need to do to work with infinite sets.
I understand that you are believing in nonsense,not mathematics.
NUF(x) being the number of unit fractions above x
No, below x. NUF(x) is the number between 0 and x.
So, it is INFINITE for all x > 0
Not according to mathematics: ∀n ∈ ℕ: 1/n - 1/(n+1) = d_n
So, what is the last discernable number?
The set is potentially infinite. But all its are instances are finite
and are followed by the infinity of dark numbers.
You CAN'T "Step Down" from omega.
It can be done. It is a general rule, forced by the well-foundedness
(see 2.12.8) of the sequence of ordinal numbers, that every strictly
decreasing sequence of ordinal numbers reaches its smallest element
after a finite number of steps. Since limit ordinals have no direct
predecessors, we have to jump down from them to some predecessor.
Ask for help elsewhere if you don't trust me.
What is 2.12.8?
It is the axiom of foundation in https://www.hs-augsburg.de/~mueckenh/Transfinity/Transfinity/pdf
Your arguemnt is assuming its conclusion. Omega isn't a Natural
Number, so you can't use arguements about Natural Numbers to talk
about it.
For steps down from omega see: https://math.stackexchange.com/questions/4626222/infinitely-deep-sets/4626228#4626228
Regards, WM
On 13.12.2023 13:36, Richard Damon wrote:
On 12/13/23 7:01 AM, WM wrote:
On 13.12.2023 00:46, Richard Damon wrote:
NUF(x) being the number of unit fractions below x is 0 for x <=0 and
infinite for x > 0, so has a major discontinuity at 0
Impossible. Contradicting mathematics: Never two or more unit
fractions sit at the same place.
Right, no two at the same space, but there "density" becomes infinite,
and thus we get the discontinuity,
No. All distances remain finite.
You just don't understand what you need to do to work with infinite sets.
I understand that you are believing in nonsense,not mathematics.
NUF(x) being the number of unit fractions above x
No, below x. NUF(x) is the number between 0 and x.
So, it is INFINITE for all x > 0
Not according to mathematics: ∀n ∈ ℕ: 1/n - 1/(n+1) = d_n
So, what is the last discernable number?
The set is potentially infinite. But all its are instances are finite
and are followed by the infinity of dark numbers.
You CAN'T "Step Down" from omega.
It can be done. It is a general rule, forced by the well-foundedness
(see 2.12.8) of the sequence of ordinal numbers, that every strictly
decreasing sequence of ordinal numbers reaches its smallest element
after a finite number of steps. Since limit ordinals have no direct
predecessors, we have to jump down from them to some predecessor.
Ask for help elsewhere if you don't trust me.
What is 2.12.8?
It is the axiom of foundation in https://www.hs-augsburg.de/~mueckenh/Transfinity/Transfinity/pdf
Your arguemnt is assuming its conclusion. Omega isn't a Natural
Number, so you can't use arguements about Natural Numbers to talk
about it.
For steps down from omega see: https://math.stackexchange.com/questions/4626222/infinitely-deep-sets/4626228#4626228
Regards, WM
On 12/13/23 11:51 AM, WM wrote:
No, below x. NUF(x) is the number between 0 and x.
So, it is INFINITE for all x > 0
Not according to mathematics: ∀n ∈ ℕ: 1/n - 1/(n+1) = d_n
Which says nothing about NUF(x)
Your arguemnt is assuming its conclusion. Omega isn't a Natural
Number, so you can't use arguements about Natural Numbers to talk
about it.
For steps down from omega see:
https://math.stackexchange.com/questions/4626222/infinitely-deep-sets/4626228#4626228
But they aren't "Natural Numbers".
Yes, we can have some "Transfinite" numbers BEYOND the Natural Numbers,
but they are not members of the Natural Numbers and do not obey all the properties of them.
On 12/13/23 11:51 AM, WM wrote:
Your arguemnt is assuming its conclusion. Omega isn't a Natural
Number, so you can't use arguements about Natural Numbers to talk
about it.
For steps down from omega see:
https://math.stackexchange.com/questions/4626222/infinitely-deep-sets/4626228#4626228
So, you are just ASSUMING a falsehood
And since Omega isn't in the set of Natural Numbers, you can't talk
about jumping from Omega to a Natural Number by this logic.
On 15.12.2023 02:54, Richard Damon wrote:
On 12/13/23 11:51 AM, WM wrote:
Your arguemnt is assuming its conclusion. Omega isn't a Natural
Number, so you can't use arguements about Natural Numbers to talk
about it.
For steps down from omega see:
https://math.stackexchange.com/questions/4626222/infinitely-deep-sets/4626228#4626228
So, you are just ASSUMING a falsehood
No, I have shown your sentence wrong:
And since Omega isn't in the set of Natural Numbers, you can't talk
about jumping from Omega to a Natural Number by this logic.
You can't understand logic. EOD.
Regards, WM
On 12/15/23 7:21 AM, WM wrote:
Yes, NUF can not increase by more than 1 without being constant for a
finite period,
but that finite period is unboundedly small,
Every step down from omega ends at a visible natural number.
But then the step up needs to get back to omega, but it doesn't
Le 16/12/2023 à 02:05, Richard Damon a écrit :
On 12/15/23 7:21 AM, WM wrote:
Yes, NUF can not increase by more than 1 without being constant for a
finite period,
Fine that you accept mathematics, after all.
but that finite period is unboundedly small,
It is finite, greater than zero, so Cantor would say it has uncountably
many points. I say: It has at least one point.
Every step down from omega ends at a visible natural number.
But then the step up needs to get back to omega, but it doesn't
You are mistaken. Ask a set theorist whom you trust.
Regards, WM
On 12/16/23 7:36 AM, WM wrote:
Le 16/12/2023 à 02:05, Richard Damon a écrit :
On 12/15/23 7:21 AM, WM wrote:
Yes, NUF can not increase by more than 1 without being constant for a
finite period,
Fine that you accept mathematics, after all.
but that finite period is unboundedly small,
It is finite, greater than zero, so Cantor would say it has uncountably
many points. I say: It has at least one point.
No, the unit fractions have a COUNTABLE infinte number of points.
And "at least one" can be a countably infinite, but doesn't imply that
there is a "first" in that sequence (since the enumeration is decreasing)
So in your mind "Step back" isn't the reversal of "Step Up"? Then what
Every step down from omega ends at a visible natural number.
But then the step up needs to get back to omega, but it doesn't
You are mistaken. Ask a set theorist whom you trust.
is it?
Le 16/12/2023 à 13:52, Richard Damon a écrit :
On 12/16/23 7:36 AM, WM wrote:
Le 16/12/2023 à 02:05, Richard Damon a écrit :
On 12/15/23 7:21 AM, WM wrote:
Yes, NUF can not increase by more than 1 without being constant for
a finite period,
Fine that you accept mathematics, after all.
but that finite period is unboundedly small,
It is finite, greater than zero, so Cantor would say it has
uncountably many points. I say: It has at least one point.
No, the unit fractions have a COUNTABLE infinte number of points.
But between two unit fractions there is at least one point.
And "at least one" can be a countably infinite, but doesn't imply that
there is a "first" in that sequence (since the enumeration is decreasing)
The enumeration is irrelevant. Relevant is only that between every two
unit fractions there is a point where the function NUF(x) does not
increase.
So in your mind "Step back" isn't the reversal of "Step Up"? Then what
Every step down from omega ends at a visible natural number.
But then the step up needs to get back to omega, but it doesn't
You are mistaken. Ask a set theorist whom you trust.
is it?
A step down from omega can only end at a visible number. But above each visisble number there are many visible numbers followed by ℵo dark
numbers which cannot be stepped to from above.
For the facts ask a set theorist.
Regards, WM
On 12/16/23 8:48 AM, WM wrote:
The enumeration is irrelevant. Relevant is only that between every two
unit fractions there is a point where the function NUF(x) does not
increase.
Yes, so BETWEEN unit fractions you have the space,
but since there is no
"lowest" unit fraction, there isn't a point that you can say it has the
value of 1.
A step down from omega can only end at a visible number. But above each
visible number there are many visible numbers followed by ℵo dark
numbers which cannot be stepped to from above.
But omega isn't a Natual Number, so you can't step down from it to one.
Omega, is a symbol that defines the size of that set, but isn't a number
in that set, so you have no basis to use the laws of Natual Numbers on it.
Le 16/12/2023 à 15:11, Richard Damon a écrit :
On 12/16/23 8:48 AM, WM wrote:
The enumeration is irrelevant. Relevant is only that between every
two unit fractions there is a point where the function NUF(x) does
not increase.
Yes, so BETWEEN unit fractions you have the space,
Between all unit fractions.
but since there is no "lowest" unit fraction, there isn't a point that
you can say it has the value of 1.
If there are all unit fractions and all have distances, then there is a
first one.
A step down from omega can only end at a visible number. But above
each visible number there are many visible numbers followed by ℵo
dark numbers which cannot be stepped to from above.
But omega isn't a Natual Number, so you can't step down from it to one.
Don't show your ignorance. Ask for help.
Omega, is a symbol that defines the size of that set, but isn't a
number in that set, so you have no basis to use the laws of Natual
Numbers on it.
For the facts ask a trusted set theorist.
Regards, WM
On 12/17/23 4:23 PM, WM wrote:
Le 16/12/2023 à 15:11, Richard Damon a écrit :
On 12/16/23 8:48 AM, WM wrote:
The enumeration is irrelevant. Relevant is only that between every
two unit fractions there is a point where the function NUF(x) does
not increase.
Yes, so BETWEEN unit fractions you have the space,
Between all unit fractions.
Right
I guess you think Achillies can't pass the tortoise,
The distances between the points shrinks as we get closer to 0, so there
is space for more and more points.
A step down from omega can only end at a visible number. But above
each visible number there are many visible numbers followed by ℵo
dark numbers which cannot be stepped to from above.
But omega isn't a Natual Number, so you can't step down from it to one.
Don't show your ignorance. Ask for help.
Really.
Omega, is a symbol that defines the size of that set, but isn't a
number in that set, so you have no basis to use the laws of Natual
Numbers on it.
For the facts ask a trusted set theorist.
Which isn't you.
Show the ACTUAL set theory that shows your claim.
Le 18/12/2023 à 01:09, Richard Damon a écrit :
On 12/17/23 4:23 PM, WM wrote:
Le 16/12/2023 à 15:11, Richard Damon a écrit :
On 12/16/23 8:48 AM, WM wrote:
The enumeration is irrelevant. Relevant is only that between every
two unit fractions there is a point where the function NUF(x) does
not increase.
Yes, so BETWEEN unit fractions you have the space,
Between all unit fractions.
Right
So NUF can not increase by more than 1 without being constant for a
finite period.
Do you uphold your claim "any distance is finite, you can have an
infinite number of them before
any finite distance"?
I guess you think Achillies can't pass the tortoise,
We can't divide the distance infinitely often because of dark numbers.
The distances between the points shrinks as we get closer to 0, so
there is space for more and more points.
But not for distinguishable points. Remember: You can have an infinite
number of them after
any finite distance.
Don't show your ignorance. Ask for help.A step down from omega can only end at a visible number. But above
each visible number there are many visible numbers followed by ℵo
dark numbers which cannot be stepped to from above.
But omega isn't a Natual Number, so you can't step down from it to one. >>>
Really.
See for instance the Goodstein-sequences described here https://www.hs-augsburg.de/~mueckenh/Transfinity/Transfinity/pdf
Omega, is a symbol that defines the size of that set, but isn't a
number in that set, so you have no basis to use the laws of Natual
Numbers on it.
For the facts ask a trusted set theorist.
Which isn't you.
Show the ACTUAL set theory that shows your claim.
See the answer in https://math.stackexchange.com/questions/3781681/well-ordering-on-natural-numbers
or see the axiom of foundation.
Regards, WM
On 12/18/23 5:40 AM, WM wrote:
Le 18/12/2023 à 01:09, Richard Damon a écrit :
On 12/17/23 4:23 PM, WM wrote:
Le 16/12/2023 à 15:11, Richard Damon a écrit :
On 12/16/23 8:48 AM, WM wrote:
The enumeration is irrelevant. Relevant is only that between every >>>>>> two unit fractions there is a point where the function NUF(x) does >>>>>> not increase.
Yes, so BETWEEN unit fractions you have the space,
Between all unit fractions.
Right
So NUF can not increase by more than 1 without being constant for a
finite period.
Right, but starts at infinity for any finite value > 0.
Do you uphold your claim "any distance is finite, you can have an
infinite number of them before
any finite distance"?
Right,
so, since there can be no "smallest finite value", the step from
0 to 1, and then to 2, and up to infinity, occurs in the infintesimal
space between 0 and finite.
It happens in the domain of unboundedly
small numbers.
I guess you think Achillies can't pass the tortoise,
We can't divide the distance infinitely often because of dark numbers.
So, when do you stop dividing?
See for instance the Goodstein-sequences described here
https://www.hs-augsburg.de/~mueckenh/Transfinity/Transfinity/pdf
IF you want a formal debugging of your theory, send me a Purchase Order authorizing payment for my time. I charge $100 per hour.
THe quick free evaluation see the use of the undefinable (or misdefined)
term "transfinite ordinal numbers" is an oxymoron, as the Ordinal
Numbers are by there definition "Finite" and don't include any infinities.
It seems you don't have much mathematics background as you seem to think
that your concept of "Transfinite" numbers is "new"
studying them over 50 years ago. As I say, you need to define your
actual construction process for them, and not just not just assume you
can extend the definition of Natural Number to include "unreachable"
numbers to get to them.
Show the ACTUAL set theory that shows your claim.
See the answer in
https://math.stackexchange.com/questions/3781681/well-ordering-on-natural-numbers
or see the axiom of foundation.
Le 18/12/2023 à 15:46, Richard Damon a écrit :
On 12/18/23 5:40 AM, WM wrote:
Le 18/12/2023 à 01:09, Richard Damon a écrit :
On 12/17/23 4:23 PM, WM wrote:
Le 16/12/2023 à 15:11, Richard Damon a écrit :
On 12/16/23 8:48 AM, WM wrote:
The enumeration is irrelevant. Relevant is only that between
every two unit fractions there is a point where the function
NUF(x) does not increase.
Yes, so BETWEEN unit fractions you have the space,
Between all unit fractions.
Right
So NUF can not increase by more than 1 without being constant for a
finite period.
Right, but starts at infinity for any finite value > 0.
It starts at 0 with value 0.
Do you uphold your claim "any distance is finite, you can have an
infinite number of them before
any finite distance"?
Right,
Then you canot distinguish them by a finite separator.
so, since there can be no "smallest finite value", the step from 0 to
1, and then to 2, and up to infinity, occurs in the infintesimal space
between 0 and finite.
Nonsense. NUF increases to 1 at the first unit fraction. It is not infinitesimal but it is dark.
It happens in the domain of unboundedly small numbers.
Come off it! All unit fractions, like all natural numbers, are finite.
I guess you think Achillies can't pass the tortoise,
We can't divide the distance infinitely often because of dark numbers.
So, when do you stop dividing?
That is potential infinity.
See for instance the Goodstein-sequences described here
https://www.hs-augsburg.de/~mueckenh/Transfinity/Transfinity/pdf
IF you want a formal debugging of your theory, send me a Purchase
Order authorizing payment for my time. I charge $100 per hour.
You are not worth 1 $ per day.
THe quick free evaluation see the use of the undefinable (or
misdefined) term "transfinite ordinal numbers" is an oxymoron, as the
Ordinal Numbers are by there definition "Finite" and don't include any
infinities.
So you have no idea of set theory? ω, ω+1, 2ω, ω^ω are transfinite ordinal numbers. See https://www.hs-augsburg.de/~mueckenh/Transfinity/Transfinity/pdf.
It seems you don't have much mathematics background as you seem to
think that your concept of "Transfinite" numbers is "new"
No it is Cantor's invention.
, but I remember
studying them over 50 years ago. As I say, you need to define your
actual construction process for them, and not just not just assume you
can extend the definition of Natural Number to include "unreachable"
numbers to get to them.
Show the ACTUAL set theory that shows your claim.
I did.
See the answer in
https://math.stackexchange.com/questions/3781681/well-ordering-on-natural-numbers
or see the axiom of foundation.
Have you understood that it is possibleto descebd from pomega?
Regards, WM
On 12/18/23 11:40 AM, WM wrote:
It starts at 0 with value 0.
And then has a discontinuity,
because it is infinite at all finite numbers > 0.
Then you canot distinguish them by a finite separator.
But I did.
Give me two finite points, and a point exists between them.
But there is no "first unit fraction",
It happens in the domain of unboundedly small numbers.
Come off it! All unit fractions, like all natural numbers, are finite.
And "Unboundedly Small" is still finite in value, just infinite in the
number of them.
It seems you are trying to define as "potential infinity" the concept of Unbounded, but note, every element of the Unbounded set is finite, but
the size of the set is infinite.
See for instance the Goodstein-sequences described here
https://www.hs-augsburg.de/~mueckenh/Transfinity/Transfinity/pdf
IF you want a formal debugging of your theory, send me a Purchase
Order authorizing payment for my time. I charge $100 per hour.
You are not worth 1 $ per day.
People pay me it.
What do people pay you for you logic?
Have you understood that it is possible to descend from omega?
yes, the number before omega is omega-1, or is not a number.
Le 18/12/2023 à 19:40, Richard Damon a écrit :
On 12/18/23 11:40 AM, WM wrote:
It starts at 0 with value 0.
And then has a discontinuity,
You contradict yourself. Yes, NUF can not increase by more than 1
without being constant for a finite period,
because it is infinite at all finite numbers > 0.
Impossible if you are right.
Then you canot distinguish them by a finite separator.
But I did.
Impossible.
Give me two finite points, and a point exists between them.
Two given points can be distingusihed. None of ℵ₀ unit fractions can be given.
But there is no "first unit fraction",
wrong.
It happens in the domain of unboundedly small numbers.
Come off it! All unit fractions, like all natural numbers, are finite.
And "Unboundedly Small" is still finite in value, just infinite in the
number of them.
Every d_n coveres uncountably many points.
It seems you are trying to define as "potential infinity" the concept
of Unbounded, but note, every element of the Unbounded set is finite,
but the size of the set is infinite.
The size is finite but can be enlarged.
EgZjaHJvbWUqCQgCEAAYExiABDIJCAAQABgTGIAEMgYIARBFGDkyCQgCEAAYExiABDIJCAMQABgTGIAEMgoIBBAAGBMYFhgeMgoIBRAAGBMYFhgeMgwIBhAAGA8YExgWGB4yDAgHEAAYDxgTGBYYHjIKCAgQABgTGBYYHjIKCAkQABgTGBYYHtIBCDYzNDdqMWo3qAIAsAIA&sourceid=chrome&ie=UTF-8
See for instance the Goodstein-sequences described here
https://www.hs-augsburg.de/~mueckenh/Transfinity/Transfinity/pdf
IF you want a formal debugging of your theory, send me a Purchase
Order authorizing payment for my time. I charge $100 per hour.
You are not worth 1 $ per day.
People pay me it.
They must be very stupid. You showed that you have not the least idea of logig when you said: "transfinite ordinal numbers" is an oxymoron. https://www.google.com/search?q=transfinite+ordinal+numbers&rlz=1C1CHBF_deDE881DE881&oq=transfinite+ordinal&gs_lcrp=
What do people pay you for you logic?
I earned enough as a professor of the THA.
Have you understood that it is possible to descend from omega?
yes, the number before omega is omega-1, or is not a number.
It is dark. Therefore you cannot enter it. You can only enter a visible number. But every number smaller than onega is a natural number.
Regards, WM
On 12/19/23 6:50 AM, WM wrote:
Le 18/12/2023 à 19:40, Richard Damon a écrit :
On 12/18/23 11:40 AM, WM wrote:
It starts at 0 with value 0.
And then has a discontinuity,
You contradict yourself. Yes, NUF can not increase by more than 1
without being constant for a finite period,
Nope, the discontinuity occurs at value outside the domain of x.
because it is infinite at all finite numbers > 0.
Impossible if you are right.
Nope, quite possible
Have you understood that it is possible to descend from omega?
yes, the number before omega is omega-1, or is not a number.
It is dark. Therefore you cannot enter it. You can only enter a visible
number. But every number smaller than onega is a natural number.
Why is it dark? Why can't I "enter" it. I gave it a name.
Note, if you have omega + 1, then you are in something beyond what
Cantor talked about, as he went from Alpha 0 to Alpha 1, with nothing in between.
In his system omega + 1 = omega
as is omega - 1 equal to omega,
so you don't have that issue.
Le 19/12/2023 à 15:28, Richard Damon a écrit :
On 12/19/23 6:50 AM, WM wrote:
Le 18/12/2023 à 19:40, Richard Damon a écrit :
On 12/18/23 11:40 AM, WM wrote:
It starts at 0 with value 0.
And then has a discontinuity,
You contradict yourself. Yes, NUF can not increase by more than 1
without being constant for a finite period,
Nope, the discontinuity occurs at value outside the domain of x.
Nonsense. There is no discontinuity. Who should cause it?
because it is infinite at all finite numbers > 0.
Impossible if you are right.
Nope, quite possible
You cannot think. It is useless to continue with this topic.
Have you understood that it is possible to descend from omega?
yes, the number before omega is omega-1, or is not a number.
It is dark. Therefore you cannot enter it. You can only enter a
visible number. But every number smaller than onega is a natural number.
Why is it dark? Why can't I "enter" it. I gave it a name.
But you cannot give the finite initial segment {1, 2, 3, ..., n}.
Note, if you have omega + 1, then you are in something beyond what
Cantor talked about, as he went from Alpha 0 to Alpha 1, with nothing
in between.
You have never learnt set theory, have you?
Alephs are cardinals. ℵo = ℵo + 1 = 2ℵo, = ℵo*ℵo. Omegas are ordinals. ω
=/= ω+1, =/= 2ω =/= ω^ω, although |ω| = |ω+1| = |2ω| = |ω^ω = ℵo.
In his system omega + 1 = omega
No!
as is omega - 1 equal to omega,
ω-1 does not exist in his system!
so you don't have that issue.
Your claims are ridiculous. But don't bother, we need no transfinite
numbers for discussing dark natural numbers.
Regards, WM
On 12/19/23 2:10 PM, WM wrote:
In his system omega + 1 = omega
No!
as is omega - 1 equal to omega,
ω-1 does not exist in his system!
So you can't step back from omega.
note, the finite and the transfinite ordinals do not "connect"
Your claims are ridiculous. But don't bother, we need no transfinite
numbers for discussing dark natural numbers.
Except that no Natural Number is dark, since all can be used individually.
Le 19/12/2023 à 22:43, Richard Damon a écrit :
On 12/19/23 2:10 PM, WM wrote:
In his system omega + 1 = omega
No!
as is omega - 1 equal to omega,
ω-1 does not exist in his system!
So you can't step back from omega.
note, the finite and the transfinite ordinals do not "connect"
We can step back from omega, but only jump to a visible natural number.
Your claims are ridiculous. But don't bother, we need no transfinite
numbers for discussing dark natural numbers.
Except that no Natural Number is dark, since all can be used
individually.
No, you can step back from omega only to visible natural numbers, which
have ℵo successors. But by using the successors collectively, you can
use them in upwards direction, like you claim that Cantor applies all
natural numbers for indexing, although every chosen number has ℵo successors.
Regards, WM
Le 20/12/2023 à 14:03, Richard Damon a écrit :
On 12/20/23 6:01 AM, WM wrote:
We can step back from omega, but only jump to a visible natural number.
From where do you get that?
Please inform yourself a bit about the foundations of set theory. I have
not in the mood to teach you.
Regards, WM
On 12/20/23 6:01 AM, WM wrote:
We can step back from omega, but only jump to a visible natural number.
From where do you get that?
On 12/20/23 11:32 AM, WM wrote:
Le 20/12/2023 à 14:03, Richard Damon a écrit :
On 12/20/23 6:01 AM, WM wrote:
We can step back from omega, but only jump to a visible natural number. >>>From where do you get that?
Please inform yourself a bit about the foundations of set theory. I am
not in the mood to teach you.
I.E, you answer isn't that you can actually show what you claim,
Le 20/12/2023 à 17:44, Richard Damon a écrit :
On 12/20/23 11:32 AM, WM wrote:
Le 20/12/2023 à 14:03, Richard Damon a écrit :
On 12/20/23 6:01 AM, WM wrote:
We can step back from omega, but only jump to a visible natural
number.
From where do you get that?
Please inform yourself a bit about the foundations of set theory. I
am not in the mood to teach you.
I.E, you answer isn't that you can actually show what you claim,
Consider as an example that you should be able to understand the sequences ω, 7, 3, 2, 1, 0
and
0, 1/4, 1/2, 1.
Do they exist? Of course. See above.
Regards, WM
On 12/22/23 2:57 AM, WM wrote:
Le 20/12/2023 à 17:44, Richard Damon a écrit :
On 12/20/23 11:32 AM, WM wrote:
Le 20/12/2023 à 14:03, Richard Damon a écrit :
On 12/20/23 6:01 AM, WM wrote:
We can step back from omega, but only jump to a visible natural
number.
From where do you get that?
Please inform yourself a bit about the foundations of set theory. I
am not in the mood to teach you.
I.E, you answer isn't that you can actually show what you claim,
Consider as an example that you should be able to understand the sequences >> ω, 7, 3, 2, 1, 0
and
0, 1/4, 1/2, 1.
Do they exist? Of course. See above.
Note, the first one is NOT a sequnce of "Natural Numbers",
And, in the same way, your second set, while it can be a set of
Rationals, it is NOT a set of the "Unit Fractions", as 0 is not a Unit Fraction.
Le 22/12/2023 à 14:59, Richard Damon a écrit :
On 12/22/23 2:57 AM, WM wrote:
Le 20/12/2023 à 17:44, Richard Damon a écrit :
On 12/20/23 11:32 AM, WM wrote:
Le 20/12/2023 à 14:03, Richard Damon a écrit :
On 12/20/23 6:01 AM, WM wrote:
We can step back from omega, but only jump to a visible natural
number.
From where do you get that?
Please inform yourself a bit about the foundations of set theory. I
am not in the mood to teach you.
I.E, you answer isn't that you can actually show what you claim,
Consider as an example that you should be able to understand the
sequences
ω, 7, 3, 2, 1, 0
and
0, 1/4, 1/2, 1.
Do they exist? Of course. See above.
Note, the first one is NOT a sequnce of "Natural Numbers",
I did not claim so.
And, in the same way, your second set, while it can be a set of
Rationals, it is NOT a set of the "Unit Fractions", as 0 is not a Unit
Fraction.
These are not sets but sequences.
Regards, WM
On 12/22/23 11:21 AM, WM wrote:
Le 22/12/2023 à 14:59, Richard Damon a écrit :
On 12/22/23 2:57 AM, WM wrote:
Le 20/12/2023 à 17:44, Richard Damon a écrit :
On 12/20/23 11:32 AM, WM wrote:
Le 20/12/2023 à 14:03, Richard Damon a écrit :
On 12/20/23 6:01 AM, WM wrote:
We can step back from omega, but only jump to a visible natural >>>>>>>> number.
From where do you get that?
Please inform yourself a bit about the foundations of set theory. I >>>>>> am not in the mood to teach you.
I.E, you answer isn't that you can actually show what you claim,
Consider as an example that you should be able to understand the
sequences
ω, 7, 3, 2, 1, 0
and
0, 1/4, 1/2, 1.
Do they exist? Of course. See above.
Note, the first one is NOT a sequnce of "Natural Numbers",
I did not claim so.
WHen I mentioned that your dark numbers could be seen as in the
transfinite numbers outside the Natual Number, you complained that all
your logic was being done inside the Natural Number, so your "Dark
Numbers" were be neccesity Natural Numbers
If 0 isn't a unit fraction, it can't be the lower bound of unit fractions.
Le 22/12/2023 à 17:55, Richard Damon a écrit :
On 12/22/23 11:21 AM, WM wrote:
Le 22/12/2023 à 14:59, Richard Damon a écrit :
On 12/22/23 2:57 AM, WM wrote:
Le 20/12/2023 à 17:44, Richard Damon a écrit :
On 12/20/23 11:32 AM, WM wrote:
Le 20/12/2023 à 14:03, Richard Damon a écrit :
On 12/20/23 6:01 AM, WM wrote:
We can step back from omega, but only jump to a visible natural >>>>>>>>> number.
From where do you get that?
Please inform yourself a bit about the foundations of set theory. >>>>>>> I am not in the mood to teach you.
I.E, you answer isn't that you can actually show what you claim,
Consider as an example that you should be able to understand the
sequences
ω, 7, 3, 2, 1, 0
and
0, 1/4, 1/2, 1.
Do they exist? Of course. See above.
Note, the first one is NOT a sequnce of "Natural Numbers",
I did not claim so.
WHen I mentioned that your dark numbers could be seen as in the
transfinite numbers outside the Natual Number, you complained that all
your logic was being done inside the Natural Number, so your "Dark
Numbers" were be neccesity Natural Numbers
Of course. And since the leap down from omega never hits a dark number,
their existence is proven by the fact that after leaping down you can
run upwards through all dark numbers collectively by "and so on" or by
"...".
If 0 isn't a unit fraction, it can't be the lower bound of unit
fractions.
Then find another one.
Regards, WM
On 12/23/23 7:40 AM, WM wrote:
Of course. And since the leap down from omega never hits a dark number,
their existence is proven by the fact that after leaping down you can
run upwards through all dark numbers collectively by "and so on" or by
"...".
But there is no "leap down". What is the leap down from 0 (in the
Naturals).
You never reach "up" from the Ordinals to the Transfinite,
If 0 isn't a unit fraction, it can't be the lower bound of unit
fractions.
Then find another one.
It doesn't exist.
There is no requirement that a bound exist, read your article on infimum:
In mathematics, the infimum (abbreviated inf; plural infima) of a subset
S of a partially ordered set P is the greatest element in P that is less
than or equal to each element of S, *if such an element exists.*
So, there need not be a bound, in fact, that is why they are called "unbounded" sets.
Le 23/12/2023 à 14:37, Richard Damon a écrit :
On 12/23/23 7:40 AM, WM wrote:
Of course. And since the leap down from omega never hits a dark
number, their existence is proven by the fact that after leaping down
you can run upwards through all dark numbers collectively by "and so
on" or by "...".
But there is no "leap down". What is the leap down from 0 (in the
Naturals).
There is a leap from omega to 17.
You never reach "up" from the Ordinals to the Transfinite,
The sequence has been given by Cantor:
1, 2, 3, ..., ω, ω+1, ω+2, ..., 2ω, 2ω+1, ...
If 0 isn't a unit fraction, it can't be the lower bound of unit
fractions.
Then find another one.
It doesn't exist.
You are so wrong that I presume you never have studied mathematics.
There is no requirement that a bound exist, read your article on infimum:
In mathematics, the infimum (abbreviated inf; plural infima) of a subset
S of a partially ordered set P is the greatest element in P that is
less than or equal to each element of S, *if such an element exists.*
So, there need not be a bound, in fact, that is why they are called
"unbounded" sets.
No, there is no bound that is an element of S. The unit fractions have a supremum that is a unit fraction, namely 1 = 1/1, and an infimum that is
not a unit fraction, namely 0.
Regards, WM
On 12/24/23 3:37 AM, WM wrote:
Le 23/12/2023 à 14:37, Richard Damon a écrit :
On 12/23/23 7:40 AM, WM wrote:
Of course. And since the leap down from omega never hits a dark
number, their existence is proven by the fact that after leaping down
you can run upwards through all dark numbers collectively by "and so
on" or by "...".
But there is no "leap down". What is the leap down from 0 (in the
Naturals).
There is a leap from omega to 17.
Really, a ONE STEP leap?
You never reach "up" from the Ordinals to the Transfinite,
The sequence has been given by Cantor:
1, 2, 3, ..., ω, ω+1, ω+2, ..., 2ω, 2ω+1, ...
Please show a reference where you can actually "Step up" from one of the
... to the following multiple of omega (or down).
BOUNDS, as I am using them are the maximum or minimum
No, there is no bound that is an element of S. The unit fractions have a
supremum that is a unit fraction, namely 1 = 1/1, and an infimum that is
not a unit fraction, namely 0.
And since the infimum isn't is the Unit fractions, there is no smallest
unit fraction, and thus no finite point where NUF(x) == 1.
Le 24/12/2023 à 18:04, Richard Damon a écrit :
On 12/24/23 3:37 AM, WM wrote:
Le 23/12/2023 à 14:37, Richard Damon a écrit :
On 12/23/23 7:40 AM, WM wrote:
Of course. And since the leap down from omega never hits a dark
number, their existence is proven by the fact that after leaping
down you can run upwards through all dark numbers collectively by
"and so on" or by "...".
But there is no "leap down". What is the leap down from 0 (in the
Naturals).
There is a leap from omega to 17.
Really, a ONE STEP leap?
Really. But we can go every number of finite steps, for instance
ω, 2023, 2022, 2021, ..., 17, 16, 15, ..., 1, 0.
At every term we can reverse the direction and can go upwards ℵo steps.
You never reach "up" from the Ordinals to the Transfinite,
The sequence has been given by Cantor:
1, 2, 3, ..., ω, ω+1, ω+2, ..., 2ω, 2ω+1, ...
Please show a reference where you can actually "Step up" from one of
the ... to the following multiple of omega (or down).
If not, then not all natural numbers can be applied in bijections.
BOUNDS, as I am using them are the maximum or minimum
That is not correct.
No, there is no bound that is an element of S. The unit fractions
have a supremum that is a unit fraction, namely 1 = 1/1, and an
infimum that is not a unit fraction, namely 0.
And since the infimum isn't is the Unit fractions, there is no
smallest unit fraction, and thus no finite point where NUF(x) == 1.
But there is a real number, namely 0, that is not reached. Hence when
going upwards from 0, the NUF(x) increases. But it cannot increase by
more than 1 before it is constat over the gap following every unit
fraction ∀n ∈ ℕ: 1/n - 1/(n+1) = d_n > 0.
Do you disagree with this mathematical formula?
Regards, WM
On 12/25/23 2:22 PM, WM wrote:
Le 24/12/2023 à 18:04, Richard Damon a écrit :
On 12/24/23 3:37 AM, WM wrote:
Le 23/12/2023 à 14:37, Richard Damon a écrit :
On 12/23/23 7:40 AM, WM wrote:
Of course. And since the leap down from omega never hits a dark
number, their existence is proven by the fact that after leaping
down you can run upwards through all dark numbers collectively by
"and so on" or by "...".
But there is no "leap down". What is the leap down from 0 (in the
Naturals).
There is a leap from omega to 17.
Really, a ONE STEP leap?
Really. But we can go every number of finite steps, for instance
ω, 2023, 2022, 2021, ..., 17, 16, 15, ..., 1, 0.
At every term we can reverse the direction and can go upwards ℵo steps.
Why is the step down from ω 2023? Why not 100000, or 1?
There is no unique value that works, so the operation isn't actually
defined.
You never reach "up" from the Ordinals to the Transfinite,
The sequence has been given by Cantor:
1, 2, 3, ..., ω, ω+1, ω+2, ..., 2ω, 2ω+1, ...
Please show a reference where you can actually "Step up" from one of
the ... to the following multiple of omega (or down).
If not, then not all natural numbers can be applied in bijections.
Why not?
The bijection rules are based only on the set 1, 2, 3, ... and don't
need the omega stuff.
BOUNDS, as I am using them are the maximum or minimum
That is not correct.
Why not?
Le 25/12/2023 à 20:41, Richard Damon a écrit :
On 12/25/23 2:22 PM, WM wrote:
Le 24/12/2023 à 18:04, Richard Damon a écrit :Why is the step down from ω 2023? Why not 100000, or 1?
On 12/24/23 3:37 AM, WM wrote:
Le 23/12/2023 à 14:37, Richard Damon a écrit :
On 12/23/23 7:40 AM, WM wrote:
Of course. And since the leap down from omega never hits a dark
number, their existence is proven by the fact that after leaping >>>>>>> down you can run upwards through all dark numbers collectively by >>>>>>> "and so on" or by "...".
But there is no "leap down". What is the leap down from 0 (in the
Naturals).
There is a leap from omega to 17.
Really, a ONE STEP leap?
Really. But we can go every number of finite steps, for instance
ω, 2023, 2022, 2021, ..., 17, 16, 15, ..., 1, 0.
At every term we can reverse the direction and can go upwards ℵo steps. >>
All that is possible. All these sequences end after a finite number of
steps. But in reverse diraction there are infinitely many step possible.
There is no unique value that works, so the operation isn't actually
defined.
You can define whateverr you like. The sequence will be finite.
You never reach "up" from the Ordinals to the Transfinite,
The sequence has been given by Cantor:
1, 2, 3, ..., ω, ω+1, ω+2, ..., 2ω, 2ω+1, ...
Please show a reference where you can actually "Step up" from one of
the ... to the following multiple of omega (or down).
If not, then not all natural numbers can be applied in bijections.
Why not?
Because all implies that none is missing.
The bijection rules are based only on the set 1, 2, 3, ... and don't
need the omega stuff.
BOUNDS, as I am using them are the maximum or minimum
That is not correct.
Why not?
If you understand German, you may consult W. Mückenheim: "Mathematik für die ersten Semester", 4th ed., De Gruyter, Berlin (2015) https://www.degruyter.com/document/doi/10.1515/9783110377347/html
Ask Wiki or a math teacher if you don't believe me.
Regards, WM
On 12/25/23 5:40 PM, WM wrote:
Why is the step down from ω 2023? Why not 100000, or 1?
All that is possible. All these sequences end after a finite number of
steps. But in reverse diraction there are infinitely many step possible.
So, your mathematics is inconsistant.
There is no unique value that works, so the operation isn't actually
defined.
You can define whatever you like. The sequence will be finite.
So, the infinite number has a finite number of predecessors. Sounds a
bit inconsistant to me.
Because all implies that none is missing.
Right, and the "all" is the ORDINARY Natural Numbers, 1, 2, 3, ...
BOUNDS, as I am using them are the maximum or minimum
That is not correct.
Why not?
If you understand German, you may consult W. Mückenheim: "Mathematik für >> die ersten Semester", 4th ed., De Gruyter, Berlin (2015)
https://www.degruyter.com/document/doi/10.1515/9783110377347/html
Ask Wiki or a math teacher if you don't believe me.
But you can't claim yourself as an athority to prove yourself.
You are just proving your "logic" is defective.
Le 26/12/2023 à 01:31, Richard Damon a écrit :
On 12/25/23 5:40 PM, WM wrote:
Why is the step down from ω 2023? Why not 100000, or 1?
All that is possible. All these sequences end after a finite number
of steps. But in reverse diraction there are infinitely many step
possible.
So, your mathematics is inconsistant.
This is set theory.
There is no unique value that works, so the operation isn't actually
defined.
You can define whatever you like. The sequence will be finite.
So, the infinite number has a finite number of predecessors. Sounds a
bit inconsistant to me.
Without dark numbers it is inconsistent.
Because all implies that none is missing.
Right, and the "all" is the ORDINARY Natural Numbers, 1, 2, 3, ...
as well as the unit fractions which *never* sit together at one and the
same place. There is no discontinuity near zero.
BOUNDS, as I am using them are the maximum or minimum
That is not correct.
Why not?
If you understand German, you may consult W. Mückenheim: "Mathematik
für die ersten Semester", 4th ed., De Gruyter, Berlin (2015)
https://www.degruyter.com/document/doi/10.1515/9783110377347/html
Ask Wiki or a math teacher if you don't believe me.
But you can't claim yourself as an athority to prove yourself.
Og course I am an authority in this question. If you don't believe, then
ask someone else. No mathematician will share your personal theory.
You are just proving your "logic" is defective.
Chuckle.
Regards, WM
On 12/26/23 5:17 AM, WM wrote:
Le 26/12/2023 à 01:31, Richard Damon a écrit :
On 12/25/23 5:40 PM, WM wrote:
Why is the step down from ω 2023? Why not 100000, or 1?
All that is possible. All these sequences end after a finite number
of steps. But in reverse diraction there are infinitely many step
possible.
So, your mathematics is inconsistant.
This is set theory.
It seems your "Set Theory" is defective.
Right, and the "all" is the ORDINARY Natural Numbers, 1, 2, 3, ...
as well as the unit fractions which *never* sit together at one and the
same place. There is no discontinuity near zero.
No one said they "sit together", just that they get unboundedly packed together.
Also, your step from 0 to some unit fraction is NEVER to the "first"
unit fraction,
Og course I am an authority in this question. If you don't believe, then
ask someone else. No mathematician will share your personal theory.
Then show an AcTUAL PROOF that isn't based on errors.
You point out that BETWEEN two unit fractions there needs to be a finite space, and then try to claim that this must apply between a point that
isn't a unit fraction.
Le 26/12/2023 à 13:44, Richard Damon a écrit :
On 12/26/23 5:17 AM, WM wrote:
Le 26/12/2023 à 01:31, Richard Damon a écrit :
On 12/25/23 5:40 PM, WM wrote:
Why is the step down from ω 2023? Why not 100000, or 1?
All that is possible. All these sequences end after a finite number
of steps. But in reverse diraction there are infinitely many step
possible.
So, your mathematics is inconsistant.
This is set theory.
It seems your "Set Theory" is defective.
It is common set theory. Obviously you don't know set theory as little
as mathematics.
Right, and the "all" is the ORDINARY Natural Numbers, 1, 2, 3, ...
as well as the unit fractions which *never* sit together at one and
the same place. There is no discontinuity near zero.
No one said they "sit together", just that they get unboundedly packed
together.
If they do not sit together at the sma epoint, then one of them is the
first leading to an increase of NUF. There is no third alternative.
Also, your step from 0 to some unit fraction is NEVER to the "first"
unit fraction,
Either one is the first or many are the first. There is no alternative.
Og course I am an authority in this question. If you don't believe,
then ask someone else. No mathematician will share your personal theory.
Then show an AcTUAL PROOF that isn't based on errors.
The infimum is by definition the greatest lower bound. There is no proof
for that definition. But there is a proof that 0 is the infimum of the
set of unit fractions because every larger x can be undercut.
You point out that BETWEEN two unit fractions there needs to be a
finite space, and then try to claim that this must apply between a
point that isn't a unit fraction.
No, it must apply every unit fraction. Therefroe there can't be two
without a gap.
Regards, WM
On 12/26/23 1:14 PM, WM wrote:
Why is the step down from ω 2023? Why not 100000, or 1?
All that is possible. All these sequences end after a finite number >>>>>> of steps. But in reverse diraction there are infinitely many step
possible.
So, your mathematics is inconsistant.
This is set theory.
It seems your "Set Theory" is defective.
It is common set theory. Obviously you don't know set theory as little
as mathematics.
I don't see it in any set theory that I know of. What statement in set
theory defines a step down from omega to be some (or ANY) finite number.
Le 27/12/2023 à 03:29, Richard Damon a écrit :
On 12/26/23 1:14 PM, WM wrote:
Why is the step down from ω 2023? Why not 100000, or 1?
All that is possible. All these sequences end after a finite
number of steps. But in reverse diraction there are infinitely
many step possible.
So, your mathematics is inconsistant.
This is set theory.
It seems your "Set Theory" is defective.
It is common set theory. Obviously you don't know set theory as
little as mathematics.
I don't see it in any set theory that I know of. What statement in set
theory defines a step down from omega to be some (or ANY) finite number.
https://en.wikipedia.org/wiki/Axiom_of_regularity https://en.wikipedia.org/wiki/Goodstein%27s_theorem
The extended proof defines P(m)(n) = f(G(m)(n), n) as follows: take the hereditary base bn representation of G(m)(n), and replace each
occurrence of the base bn with the first infinite ordinal number ω.
Regards, WM
On 12/27/23 3:47 AM, WM wrote:
Which is against the laws of numbers,I don't see it in any set theory that I know of. What statement in set
theory defines a step down from omega to be some (or ANY) finite number.
https://en.wikipedia.org/wiki/Axiom_of_regularity
https://en.wikipedia.org/wiki/Goodstein%27s_theorem
The extended proof defines P(m)(n) = f(G(m)(n), n) as follows: take the
hereditary base bn representation of G(m)(n), and replace each
occurrence of the base bn with the first infinite ordinal number ω.
Le 27/12/2023 à 14:37, Richard Damon a écrit :
On 12/27/23 3:47 AM, WM wrote:
Which is against the laws of numbers,I don't see it in any set theory that I know of. What statement in
set theory defines a step down from omega to be some (or ANY) finite
number.
https://en.wikipedia.org/wiki/Axiom_of_regularity
https://en.wikipedia.org/wiki/Goodstein%27s_theorem
The extended proof defines P(m)(n) = f(G(m)(n), n) as follows: take
the hereditary base bn representation of G(m)(n), and replace each
occurrence of the base bn with the first infinite ordinal number ω.
If you are unable to read and understand the given references, then you
are unable to discuss this topic.
EOD.
Regards, WM
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